concept-development-studies-in-chemistry-2013-1.7.pdf

Concept Development Studies in Chemistry 2013

By: John S. Hutchinson

Concept Development Studies in Chemistry 2013

By: John S. Hutchinson

Online: < http://cnx.org/content/col11579/1.1/ >

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This selection and arrangement of content as a collection is copyrighted by John S. Hutchinson. It is licensed under the Creative Commons Attribution License 3.0 (http://creativecommons.org/licenses/by/3.0/). Collection structure revised: October 7, 2013 PDF generated: July 14, 2015 For copyright and attribution information for the modules contained in this collection, see p. 256.

Table of Contents 1 Atomic Molecular Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 Atomic Masses and Molecular Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3 Structure of an Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4 Electron Shell Model of an Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 31 5 Quantum Electron Energy Levels in an Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 39 6 Electron Orbitals and Electron Congurations in Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 7 Covalent Bonding and Electron Pair Sharing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 8 Molecular Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9 Energy and Polarity of Covalent Chemical Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 10 Bonding in Metals and Metal-Non-Metal Salts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 11 Molecular Geometry and Electron Domain Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 12 Measuring Energy Changes in Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13 Reaction Energy and Bond Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14 Physical Properties of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 139 15 The Kinetic Molecular Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 16 Phase Transitions and Phase Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 161 17 Phase Equilibrium and Intermolecular Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 18 Phase Equilibrium in Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 183 19 Reaction Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 20 Reaction Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 21 Reaction Equilibrium in the Gas Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 22 Acid Ionization Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 23 Entropy and the Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 24 Free Energy and Thermodynamic Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Attributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .256

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Chapter 1

Atomic Molecular Theory

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1.1 Introduction In this study, our goal is to develop the concept of the Atomic Molecular Theory. This is the theory at the foundation of everything we understand about Chemistry, as it states that all matter is made up of individual particles called atoms, which combine in ways that are both simple and complex to form larger particles called molecules. When we understand these atoms and molecules, it changes the way that we look at the world around us. We can understand the properties of the substances we interact with, we can make predictions about the changes and reactions that these substances will undergo, and we can design materials with properties that would be useful to us. The idea that everything is made of atoms is something we are told at a very early age, and for many students, it is hard then to imagine a world in which we don't know that everything is made of atoms. On the other hand, this particulate view of matter does seem counter to almost all of our own observations. The desks in front of us, the air we breathe, the water we drink, and even our own esh show no signs of these particles. Quite the opposite: they seem to be either very solid or quite uid, and certainly not grainy like a collection of particles might be expected to be. In this concept development study, then, we set aside our knowledge of these atoms and molecules and ask, quite skeptically, why do we believe that there are atoms that combine to form molecules? Or asked another way, if we believe that all matter is made up of atoms, how would we show that this is true? What is the evidence? Does the proof require us to see atoms, or is it possible to prove that they exist without actually seeing them?

1.2 Foundation Chemistry is the study of matter, so it makes sense for us to agree on what we mean by matter and what we want to know about it. Technically, matter is anything that has mass, but more commonly, matter is what we regard as stu. Anything that has physical properties and takes up space, whether a solid, a liquid, or gas, is matter. Matter can be anything from microscopic to galactic, or from rocks and air to butteries and humans. But we can go further than this and focus on a specic type of matter called a pure substance. This is a material that is completely uniform in properties regardless of the size of the sample we take or from where we take the sample. It is easiest to understand a pure substance by comparing it to a mixture, which may or may not be uniform in its properties such as color, density, and texture and can vary depending on how we make the mixture or its origin. Showing that a substance is either a pure substance or a mixture requires a lot of experimentation, but we will assume for our foundation that we have already identied which samples of matter are pure substances and which are mixtures. 1 This content is available online at .

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CHAPTER 1. ATOMIC MOLECULAR THEORY

As of 2013, the most comprehensive list of chemical substances numbered over 70 million entries. This huge number of materials seems incomprehensible, far beyond our understanding. However, it turns out that these 70 million substances are all made up from a much smaller set of pure substances called elements. An element is a substance, which cannot be broken down into simpler substances. There are only about 90 commonly occurring elements on earth. The remaining 70 million pure substances are combinations of these elements called compounds, and these are distinguishable from the elements in that a compound can be broken down into the elements from which it is made. For example, metallic iron and gaseous oxygen are both elements, which cannot be reduced into simpler substances, but common iron rust, ferrous oxide, is a compound made up from iron and oxygen. Therefore, rust can be reduced to iron and oxygen, and rust can be created by combining iron and oxygen. But iron and oxygen are elements; they cannot be transformed into one another and are not composed of simpler or common materials. Determining whether a pure substance is an element or a compound is a dicult and time-consuming process of experimentation. We will assume for our study that the elements have all been identied. Since matter is anything that can have mass, we will spend much of our time in this study analyzing mass. Without proving it, we will assume the validity of the Law of Conservation of Mass, an experimental result that simply says, The total mass of all products of a chemical reaction is equal to the total mass of all reactants of the reaction. In other words, matter cannot be created or destroyed by chemical or physical processes. This law makes it possible for us to measure masses of materials during reactions knowing that these masses aren't variable or unpredictable. With these assumptions in mind, we can proceed directly to experiments which led to the development of the Atomic Molecular Theory.

1.3 Observation 1: Mass Relationships during Chemical Reactions Since matter is anything that has mass, then the Law of Conservation of Mass suggests that matter is also conserved during chemical reactions: whatever we start with, we wind up with, at least in total. However, this does not mean that matter must be made up of atoms. It simply says that matter is reorganized in some way to produce new substances with new properties when a reaction takes place. Since we know that all substances are made of elements, then we can analyze the masses of the elements that participate in a chemical reaction. Most importantly, we can take a compound, break it into the elements that it is made of, and then nd the masses of those elements. From the Law of Conservation of Mass, the total mass of the elements that make up the compound must equal the mass of the starting compound. For example, a particular compound called copper carbonate is composed of the elements copper, carbon, and oxygen. If we take a 100.0 g sample of copper carbonate, we nd that it contains 51.5 g of copper, 38.8 g of oxygen, and 9.7 g of carbon. The total of these three masses is 51.5 g+38.8 g+9.7 g = 100.0g and is the same as the mass of the copper carbonate. This turns out to always be true. It does not matter what sample of copper carbonate we analyze, where it came from, who gave it to us, or how we made it. We always get these same masses of the element. What if we take 200.0 g of copper carbonate instead? Experiments show us that we get 103.0 g of copper, 77.6 g of oxygen, and 19.4 g of carbon. The total of these three masses is 103.0 g+77.6 g+19.4 g = 200.0g, so again the total mass is conserved. Even more importantly, there is something striking about these numbers when compared to the 100.0 g sample. When we double the mass of the copper carbonate, we also double the amount of copper it contains: 200.0 g of copper carbonate contains 103.0 g of copper, and 100.0 g of copper carbonate contains 51.5 g. The same is true for the amounts of oxygen and carbon. One way to look at this is that the fraction of the copper carbonate which is copper is the same in both samples: 51.5 g/100.0 g is equal to 103.0 g/200.0 g, which is equal to 51.5%. This is a very important result. After looking at data from many experiments, we nd that regardless of the specic sample of copper carbonate, regardless of the mass of that sample, and regardless of where that sample came from, the fraction of the mass of the sample which is copper is always the same, 51.5%. We get similar results for the oxygen and carbon. The fractions of the mass of every sample of copper carbonate are always 51.5% copper, 38.8% oxygen, and 9.7% carbon. Available for free at Connexions

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Figure 1.1: A sample of copper carbonate basic, CuCO

Figure

1.2:

A

graph

showing

ratios

of

3 ·Cu(OH)2

components

in

illustrating the blue-green color.

copper

2

carbonate

(photograph

from

http://woelen.homescience.net/science/chem/compounds/index.html ).

Other compounds show similar results. For example, every sample of the compound lead sulde contains 86.7% lead and 13.3% sulfur by mass. This is true whether we take 1.00 g of lead sulde or 1.00 kg of lead sulde or any other total mass. We always get the same proportions of the masses of lead and sulfur in the sample. 2 http://woelen.homescience.net/science/chem/compounds/index.html

Available for free at Connexions

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CHAPTER 1. ATOMIC MOLECULAR THEORY

This experimental observation is so consistent for the vast majority of all compounds that we regard it as a natural law, a summary of many, many observations that we therefore always expect to observe in future observations. In this case, the natural law we have observed is the Law of Denite Proportions: Law of Denite Proportions: When two or more elements combine to form a compound, their masses in that compound are in a xed and denite ratio. This means that if we break a compound down into its elements and measure the masses of the elements that make it up, those masses are always in the same ratio. We can illustrate this with a set of simple compounds each containing two of the elements of hydrogen, nitrogen, and oxygen. The denite ratios given by the Law of Denite Proportions are show in Table 1.1: Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen for 100.0 g of each compound:

Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen Compound Water Ammonia Nitric Oxide

Total Mass (g) 100.0 100.0 100.0

Mass of Hydrogen (g) 11.2 17.7 -

Mass of Nitrogen (g) 82.3 46.7

Mass of Oxygen (g) 88.8 53.3

Table 1.1

Do these xed masses mean that we are combining tiny particles of xed mass together to form these compounds? To test this hypothesis, let's look at these numbers more closely to see if we can make any sense of them. One way to look at these data is to imagine taking a sample of water that contains exactly 1.00 g of hydrogen, and also a sample of ammonia that contains exactly 1.00 g of hydrogen. If we do this, the data now look as shown in the rst two lines of Table 1.2: Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen.

Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen Compound Water Ammonia Nitric Oxide

Total Mass (g) 8.93 5.65 2.14

Mass of Hydrogen (g) 1.00 1.00 -

Mass of Nitrogen (g) 4.65 1.00

Mass of Oxygen (g) 7.93 1.14

Table 1.2

Looking at the data this way allows us to compare how a xed amount of hydrogen combines with either nitrogen or oxygen. From the Law of Denite Proportions, we know we will always get the xed mass ratios shown in Table 1.2: Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen. We might imagine that this means that water is formed from 1 atom of hydrogen and 1 atom of oxygen, and that therefore the mass of 1 atom of oxygen is 7.93 times greater than the mass of 1 atom of hydrogen. If this is true, then ammonia might also be formed from 1 atom of hydrogen and 1 atom of nitrogen, in which case 1 atom of nitrogen has a mass 4.65 times greater than 1 atom of hydrogen. This all seems consistent with the idea that these elements are made up of atoms combining in one-to-one ratios. But now we nd a problem: if an atom of oxygen is 7.93 times as massive as an atom of hydrogen and if an atom of nitrogen is 4.65 times as massive as an atom of hydrogen, then it must be that an atom of oxygen is more massive than an atom of nitrogen by the ratio of 7.93/4.65 = 1.70. When we now look at the third row of Table 1.2: Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen, the data tell us if nitric oxide is made up 1 atom of nitrogen and 1 atom of oxygen, an oxygen atom has a Available for free at Connexions

5 mass 1.14 times greater than the mass of a nitrogen. Our numbers and our conclusion aren't consistent with the experimental data, so we must have made an incorrect assumption. One possibility is that we were wrong when we assumed that there are atoms of the three elements combining to form these three compounds. But this does not seem likely, since it is hard to understand the xed mass proportions without thinking that we are combining particles with xed mass proportions. Still, we must have made an incorrect assumption since our conclusions were contradictory. Recall that in doing our calculations of the masses of the atoms, we assumed that in each compound one atom of each element combined with one atom of the other element. Although this is a simple assumption, there is no reason why only one atom of each type might combine. Perhaps the ratios are dierent than this, and atoms combine in ratios of 1-to-2, 2-to-3, or any other simple combination. The problem that this poses is that we don't have a way to proceed from here. Even if we assume that the Law of Denite Proportions tells us that the elements are made up of atoms, we have no way to determine anything about these atoms. Without knowing the ratios of atoms in dierent compounds, we cannot determine the masses of the atoms of the elements. And without knowing the masses of the atoms of the elements, we cannot determine the ratios of the atoms in dierent compounds. Without knowing anything about the atoms of these elements, we do not have a basis for believing that these elements are made up of atoms. The Atomic Molecular Theory is still outside our reach. Without further observations, we cannot say for certain whether matter is composed of atoms or not.

1.4 Observation 2: Multiple Mass Ratios We discovered above that we cannot conclude from the Law of Denite Proportions how many atoms of each element combine to form a particular compound, even if we assume that this is how a compound is formed. There is additional evidence that the Law of Denite Proportions is not nal proof of the existence of atoms. It is easy to nd dierent compounds with dierent chemical and physical properties, which are formed from the same two elements. This means that, if there are atoms, they can combine in many dierent ways. For example, there are a huge number of simple hydrocarbon compounds formed just from hydrogen and carbon. Table 1.3: Mass Relationships of Simple Compounds of Hydrogen and Carbon lists the mass percentages of just a few of these:

Mass Relationships of Simple Compounds of Hydrogen and Carbon Compound Methane Ethane Benzene

Total Mass(g) 100.0 100.0 100.0

Mass of Hydrogen(g) 25.1 20.1 7.7

Mass of Carbon(g) 74.9 79.9 92.3

Table 1.3

The Law of Denite Proportions says that the ratio of the masses of two elements in a compound is xed. In Table 1.3: Mass Relationships of Simple Compounds of Hydrogen and Carbon we see several ratios of the masses of carbon and hydrogen. This is consistent with the Law of Denite Proportions, though, because each xed ratio gives us a dierent compound with dierent chemical and physical properties. Therefore, dierent proportions of elements are possible, and depending on the elements, there may be one or many possible proportions and compounds. This is referred to as multiple proportions. Each compound gives us one denite proportion of the elements, but because there can be many compounds, there can be multiple dierent but denite proportions. Table 1.3: Mass Relationships of Simple Compounds of Hydrogen and Carbon provides just three examples of compounds formed from carbon and hydrogen. The number of such compounds is huge, each with a dierent denite mass ratio and each with its own distinct physical Available for free at Connexions

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CHAPTER 1. ATOMIC MOLECULAR THEORY

and chemical properties. For example, methane gas is commonly burned in gas stoves and liquid octane is commonly used in cars; yet both are compounds of carbon and hydrogen only. We will now look in great detail at a few compounds formed just from nitrogen and oxygen, simply called nitrogen oxides. Since we don't know anything about these compounds, for now we'll just call them Oxide A, Oxide B, and Oxide C. These three compounds are very dierent from one another. Two of these are quite toxic, but one is used as an anesthetic, particularly by dentists. Two of these are colorless, but one is a dark brown color. Let's look at the mass ratios for these three compounds in Table 1.4: Mass Relationships of Simple Compounds of Nitrogen and Oxygen.

Mass Relationships of Simple Compounds of Nitrogen and Oxygen Compound Oxide A Oxide B Oxide C

Total Mass(g) 100.0 100.0 100.0

Mass of Nitrogen(g) 30.5 46.7 63.7

Mass of Oxygen(g) 69.5 53.3 36.3

Table 1.4

At rst glance, there is nothing special about these numbers. There are no obvious patterns or relationships amongst the masses or mass ratios. But let's look at this data in the same way as we did in Table 1.2: Mass Relationships of Simple Compounds of Hydrogen, Nitrogen and Oxygen by nding the mass of oxygen that combines with 1.00 g of nitrogen. This is in Table 1.5: Mass Relationships of Simple Compounds of Nitrogen and Oxygen.

Mass Relationships of Simple Compounds of Nitrogen and Oxygen Compound Oxide A Oxide B Oxide C

Total Mass(g) 3.28 2.14 1.57

Mass of Nitrogen(g) 1.00 1.00 1.00

Mass of Oxygen(g) 2.28 1.14 0.57

Table 1.5

You might have to look at these data very hard to see it, but there is a pattern that is obvious once you see it. In the column for the Mass of Oxygen, the three values listed have a simple relationship: each one is a multiple of 0.57. We can see this most clearly if we divide each of the masses for the three oxides by 0.57. This shows us that the ratio 2.28 : 1.14 : 0.57 is equal to the ratio 4 : 2 : 1. What does this tell us? It means that if we have a xed mass of nitrogen, the mass of oxygen which will combine with it cannot be simply any amount. In fact, the opposite is true. There are a few specic masses of oxygen which will combine with the xed nitrogen, and those specic masses are integer multiples of a xed unit of mass. It is particularly interesting that the masses of oxygen are in integer ratios. Integers are a special set of numbers used for one primary purpose, which is to count objects. In this case, the object must be a xed unit of mass of oxygen. The data in Table 1.5: Mass Relationships of Simple Compounds of Nitrogen and Oxygen tell us that when we have a xed amount of nitrogen, it can be combined only with some integer number of a xed unit of mass of oxygen. Why would there be a xed unit of mass of oxygen? The simplest and best explanation is that oxygen exists as xed units of mass, or particles, and we call these particles atoms of oxygen. Thus, the data in Table 1.5: Mass Relationships of Simple Compounds of Nitrogen and Oxygen lead us to a conclusion that the element oxygen is composed of individual atoms with identical mass. We have shown that matter is made up of particles and that elements consist of identical particles or atoms. Available for free at Connexions

7 We can see these simple integer ratios in other compounds, as well. Let's look back at Table 1.3: Mass Relationships of Simple Compounds of Hydrogen and Carbon, which shows compounds of carbon and hydrogen. The ratios of the masses don't appear to be interesting until we do the same type of analysis that we did on the nitrogen oxides. Let's x the mass of hydrogen in each of these compounds and nd out the masses of carbon. The results are in Table 1.6: Mass Relationships of Simple Compounds of Hydrogen and Carbon.

Mass Relationships of Simple Compounds of Hydrogen and Carbon Compound Methane Ethane Benzene

Total Mass(g) 3.98 4.97 12.99

Mass of Hydrogen(g) 1.00 1.00 1.00

Mass of Carbon(g) 2.98 3.97 11.99

Table 1.6

The ratio 2.98 : 3.97 : 11.99 is the same as the ratio 3 : 4 : 12. Once again for a xed amount of hydrogen, the mass of carbon that can combine is a multiple of a xed mass unit. Therefore, carbon can only react in small xed units of mass, so carbon is made up of atoms. These data also show the same conclusion for hydrogen. All we have to do is x the carbon mass and compare the masses of hydrogen that will combine with that amount of carbon. The observations we have made for carbon and hydrogen compounds and nitrogen and oxygen compounds are general. They apply to all simple compounds. Thus, we state a new natural law which summarizes these observations, the Law of Multiple Proportions: Law of Multiple Proportions: When two elements combine to form more than one compound, the masses of one element that combine with a xed mass of the other element are in simple integer ratio. The Law of Multiple Proportions is rather wordy and dicult to state, but the data in Table 1.5: Mass Relationships of Simple Compounds of Nitrogen and Oxygen and Table 1.6: Mass Relationships of Simple Compounds of Hydrogen and Carbon illustrate it clearly. If we x the mass of one element, the masses of the other element in the three compounds are always in a simple integer ratio. Since these observations are general, then the conclusions that follow are also general: all elements are made up of xed units of mass, and we call these particles atoms. These conclusions point to other very important conclusions, as well. Since compounds are formed from the elements, then compounds consist of atoms of the elements in various combinations. Very importantly, the integer ratios of the masses of the atoms are always simple, that is, small integers. Therefore, the atoms of the dierent elements combine in simple ratios, too. This means that the small particles called atoms are combining in simple ways to form small particles of the compound. We call these particles molecules, and compounds consist of identical molecules made up of atoms in simple integer ratios. The Atomic Molecular Theory: We have now observed experimental data which reveal to us several important conclusions. We call these conclusions postulates, and taken together these postulates form the Atomic Molecular Theory: 1. Each element is composed of very small, identical particles called atoms. 2. All atoms of a single element have the same characteristic mass. 3. The number and masses of these atoms do not change during a chemical transformation. 4. Each compound consists of identical molecules, which are small, identical particles formed of atoms combined in simple whole number ratios. At this point, we know very little about these atoms, other than that they exist, have xed mass for each element, and combine to form molecules. We don't know what the atomic masses are, even in comparison to each other, and we don't know the simple integer ratios by which they combine to form molecules and compounds. We have made a major step forward in proving that matter is made up of atoms and molecules, but we have a long way to go to make this theory useful. Available for free at Connexions

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CHAPTER 1. ATOMIC MOLECULAR THEORY

1.5 Review and Discussion Questions 1. Assume that matter does not consist of atoms. Show by example how this assumption leads to hypothetical predictions which contradict the Law of Multiple Proportions. Do these hypothetical examples contradict the Law of Denite Proportions? Are both observations required for conrmation of the atomic theory? 2. Two compounds, A and B, are formed entirely from hydrogen and carbon. Compound A is 80.0% carbon by mass, and 20.0% hydrogen, whereas Compound B is 83.3% carbon by mass and 16.7% hydrogen. Demonstrate that these two compounds obey the Law of Multiple Proportions. Explain why these results strongly indicate that the elements carbon and hydrogen are composed of atoms. 3. In many chemical reactions, mass does not appear to be a conserved quantity. For example, when a tin can rusts, the resultant rusty tin can has a greater mass than before rusting. When a candle burns, the remaining candle has invariably less mass than before it was burned. Provide an explanation of these observations, and describe an experiment which would demonstrate that mass is actually conserved in these chemical reactions. 4. The following question was posed on an exam: An unknown non-metal element (Q) forms two gaseous

uorides of unknown molecular formula. A 3.2 g sample of Q reacts with uorine to form 10.8 g of the unknown uoride A. A 6.4 g sample of Q reacts with uorine to form 29.2 g of unknown uoride B. Using these data only, demonstrate by calculation and explanation that these unknown compounds obey the Law of Multiple Proportions. A student responded with the following answer:

The Law of Multiple Proportions states that when two elements form two or more compounds, the ratios of the masses of the elements between the two compounds are in a simple whole number ratio. So, looking at the data above, we see that the ratio of the mass of element Q in compound A to the mass of element Q in compound B is 3.2:6.4 = 1:2, which is a simple whole number ratio. This demonstrates that these compounds obey the Law of Multiple Proportions.

Assess the accuracy of the students answer. In your assessment, you must determine what information is correct or incorrect, provide the correct information where needed, explain whether the reasoning is logical or not, and provide logical reasoning where needed. By John S. Hutchinson, Rice University, 2011

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Chapter 2

Atomic Masses and Molecular Formulas

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2.1 Introduction In this study, our goals are to determine the masses of the atoms of each element and to nd the ratios of the atoms that combine to form the molecules of dierent compounds. We will nd that we determine the atomic masses relative to one another. In other words, we will nd that the atoms of one element may be 1.2 times as massive as the atoms of another element. We will not actually determine the mass of an atom relative to, say, a large sample of matter like a bar of metal or a glass of water. For almost all purposes in Chemistry, it turns out that we only need these relative masses. We will also nd the formulas of individual molecules for dierent compounds. The molecular formula tells us how many atoms of each type there are in a molecule of a compound. For example, many people know that water is H2 O, meaning that a molecule of water contains two hydrogen atoms and one oxygen atom. Although we haven't shown it yet, the molecular formula of water is H2 O. The postulates of the Atomic Molecular Theory provide us a great deal of understanding of pure substances and chemical reactions. For example, the theory reveals the distinction between an element and a compound. In an element, all atoms are identical. In a compound, there are atoms of two or more elements combined into small identical molecules in small integer ratios. The theory also reveals to us what happens during a chemical reaction. When two elements react, their atoms combine to form molecules in xed ratios, making a new compound. When two compounds react, the atoms in the molecules of these reactant compounds recombine into new molecules of new product compounds. As good as this is, it is about as far as we can go without further observations and analysis. We don't know the relative masses of the atoms, and we don't know the molecular formula for any compound. It turns out that to know one of these things we need to know the other one. To see this, let's take another look at the data in the Concept Development Study on Atomic Molecular Theory (Table 1.5: Mass Relationships of Simple Compounds of Nitrogen and Oxygen). Here it is again as Table 2.1: Mass Relationships of Simple Compounds of Nitrogen and Oxygen:

Mass Relationships of Simple Compounds of Nitrogen and Oxygen Compound Oxide A Oxide B Oxide C

Total Mass (g) 3.28 2.14 1.57

Mass of Nitrogen (g) 1.00 1.00 1.00

Mass of Oxygen (g) 2.28 1.14 0.57

Table 2.1

1 This content is available online at . Available for free at Connexions

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CHAPTER 2. ATOMIC MASSES AND MOLECULAR FORMULAS

We now know that a xed mass of nitrogen means a xed number of nitrogen atoms, since each atom has the same mass. And if we double the mass of oxygen, we have doubled the number of oxygen atoms. So, comparing Oxide A to Oxide B, for the same number of nitrogen atoms, Oxide A has twice as many oxygen atoms as Oxide B, which has twice as many oxygen atoms as Oxide C. There are many possible molecular formulas which are consistent with these ratios. To see this, we can show that any one of the oxides A, B, or C could have the molecular formula NO. If Oxide C has the molecular formula NO, then Oxide B has the formula NO2 , and Oxide A has the formula NO4 . If Oxide B has molecular formula NO, then Oxide A has formula NO2 , and Oxide C has formula N2 O. It might not be clear that Oxide C would be N2 O. The mass data tell us that Oxide B has twice as many oxygen atoms per nitrogen atom as Oxide C. So if Oxide B and Oxide C have the same number of oxygen atoms, then Oxide C has twice as many nitrogen atoms as Oxide B. As a similar example, if Oxide A has formula NO, then Oxide B has formula N2 O and Oxide C has formula N4 O. These three possibilities are listed in Table 2.2: Possible Molecular Formulae for Nitrogen Oxides.

Possible Molecular Formulae for Nitrogen Oxides

Assuming that: Oxide A is Oxide B is Oxide C is

Oxide C is NO NO4 NO2 NO

Oxide B is NO NO2 NO N2 O

Oxide A is NO NO N2 O N4 O

Table 2.2

We don't have a way to know which of these sets of molecular formulas are right, since all three sets are consistent with the data we have. How can we pick the right one? If we had some way to count the numbers of atoms in a sample of each compounds, then we would know. This sounds quite dicult, though. On the other hand, if we knew that the ratio of the mass of an nitrogen atom to an oxygen atom is 2.28:1.00, which is the mass ratio in Oxide A, then we could know that Oxide A is NO. But we don't have a way to take the mass of an individual atom, even on a relative basis. What we have learned is that, if we know the relative masses of the atoms, we can determine molecular formulas. And if we know the molecular formulas, we can determine relative atomic masses. We need one or the other to move forward.

2.2 Foundation We are assuming that we know the postulates of the Atomic Molecular Theory, as developed in the rst Concept Development Study. These are: (1) the elements are comprised of identical atoms; (2) all atoms of a single element have the same characteristic mass; (3) the number and masses of these atoms do not change during a chemical transformation; and (4) compounds consist of identical molecules formed of atoms combined in simple whole number ratios. We will base much of our work on the observed natural laws on which our theory is based: the Law of Conservation of Mass, the Law of Denite Proportions, and the Law of Multiple Proportions. We will be making observations about the physical properties of gas samples, particularly the volumes of gases measured under conditions with a xed temperature and a xed pressure. We will not need much technical information about temperature and pressure. For now, we will simply stick with the common understanding that temperature is a measure of how hot or cold a sample of a substance is. Temperature can be measured by a thermometer, which is any kind of gadget which gives us the same value for two objects that are in contact with each other so that they have the same temperature. Pressure is a measure of what is sometimes called the spring of air, which is the force with which a gas resists compression. There are a number of ways to measure pressure, but we will only need to know that we have a way to take measurements on gas samples such that they have the same pressure. Available for free at Connexions

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2.3 Observation 1: Volumes of Gases during Chemical Reactions During chemical reactions, some chemical and physical properties such as the total mass of the materials remain unchanged, but most properties do change. We commonly observe changes in properties when new materials are made. For example, products of reactions in comparison to the reactants may appear harder or softer, more or less colorful, more or less brittle, and more or less dense. For gases that react, volume is one of those properties that is not always conserved. A famous explosive reaction of gases involves the burning of hydrogen gas in oxygen gas to form water vapor, as shown in Figures 1 and 2. If 1 liter of oxygen gas reacts with 2 liters of hydrogen gas, the product water vapor will occupy 2 liters with no hydrogen or oxygen gas left over. (This is true if the volumes are measured with all gases at the same temperature and pressure.) Notice that the total volume of gases is not conserved: the combined volume of the reactants is 3 liters, but the volume of the product is 2 liters.

Figure 2.1:

Demonstration of an exploding balloon containing hydrogen and oxygen done at Rice

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University by Dr. Mary McHale (http://chemistry.rice.edu/ ).

2 http://chemistry.rice.edu/

Available for free at Connexions

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CHAPTER 2. ATOMIC MASSES AND MOLECULAR FORMULAS

Figure

2.2:

The

Hindenburg,

3

a

German

passenger

airship,

bursting

into

ames

in

1937

(http://www.nlhs.com/tragedy.htm ).

There is, however, something striking about the data just given for reaction of hydrogen and oxygen that should cause us to take a closer look the property of volume: the volumes of the gases are in simple integer ratios. This might look like a simple consequence of our choosing to react 1 liter of oxygen with 2 liters of hydrogen. But we can take any volumes we want, not just integer number of liters. We can react 1.42 liters of oxygen with 2.84 liters of hydrogen, and all of the hydrogen and oxygen will be consume to form 2.84 liters of water vapor. Notice that the ratio of volumes is oxygen:hydrogen:water vapor = 1:2:2, a simple integer ratio. If we try to react a dierent ratio, like 1.42 liters of oxygen with 3.00 liters of hydrogen, there will be leftover hydrogen when the reaction is complete. This is one way of observing the Law of Denite Proportions. We cannot choose to react arbitrary amounts of hydrogen and oxygen, since they combine in a xed ratio by mass. We thus observe that the volumes of hydrogen and oxygen that react, as well as the volume of water vapor product, are in a simple integer ratio. The only requirement for this to be true is that all of the gas volumes are measured with the same temperature and pressure. This result is quite general when we observe chemical reactions involving gases. The volume of hydrogen that will react with 1.0 L (liters) of nitrogen is 3.0 L, and the product of the reaction ammonia gas has volume 2.0 L. The ratio of volumes is a simple set of integers. Hydrogen chloride gas is formed from reacting hydrogen gas with chlorine gas, and again, 1.0 L of chlorine will only react with 1.0 L of hydrogen gas, with no hydrogen or chlorine left over and with 2.0 L of hydrogen chloride gas produced. Many, many such observations can be made leading us to a general law of nature called the Law of Combining Volumes: Law of Combining Volumes: When gases combine during a chemical reaction at xed temperature and pressure, the volumes of the reacting gases and products are in simple integer ratios. 3 http://www.nlhs.com/tragedy.htm

Available for free at Connexions

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2.4 Avogadro's Law  Counting Particles The integer ratios in the Law of Combining Volumes are very striking. There are few quantities in nature which are measured in integers, so it is always surprising and revealing to discover integers in measurements. Integers are generally measured only when we are counting particles or taking ratios of particles. Observing volumes in simple integer ratios should be very revealing. Looking back at the data for hydrogen and oxygen, we see that the volumes are in a simple 2:1 ratio. One of our major conclusions expressed in the Atomic Molecular Theory is that atoms and molecules react in simple integer ratios. A possible explanation of the integer volume ratios is that these are the same integer ratios as the particles react in. In the hydrogen-oxygen case, this means that the ratio of hydrogen atoms to oxygen atoms reacting is 2:1, since the volume ratio is 2:1. If the volumes are in a 2:1 ratio and the particles are in a 2:1 ratio, then a powerful conclusion emerges: equal volumes of the two gases must contain equal numbers of particles, regardless of whether they are hydrogen or oxygen. This seems like quite a leap to make, since we are concluding something about the numbers of particles without ever having counted them. What is the basis for this leap of logic? The most important part of the reasoning is the uniqueness of the integers. It is hard to come up with a simple explanation for why gas volumes should only react in integer ratios. The only simple conclusion is the one we have come to and that was rst stated by Avogadro: Avogadro's Law: Equal volumes of gas contain equal numbers of particles, if the volumes are measured at the same temperature and pressure. One way to come to this conclusion is to imagine that it is not true. Assume that equal volumes of gases do not contain equal numbers of particles and instead contain unrelated numbers of particles. Let's assume for example that 1 L of gas A contains 3.14 times as many particles as 1 L of gas B. Then to take equal numbers of A and B particles, we would need 3.14 L of gas B for every 1 L of gas A. For the gas particles of A and B to react in a simple integer ratio of particles, we would then need a non-integer ratio of volumes. But this is not what is observed in the Law of Combining Volumes: the volume of A and B that react are always observed to be a simple integer ratio. Our assumption that equal volumes contain unrelated numbers of particles leads us to a conclusion that is contradicted by experiments, so our assumption must be wrong. Therefore, equal volumes of gases contain equal numbers of particles. We can conclude that Avogadro's Law follows logically from the Law of Combining Volumes. There is a problem that we have to work out. Looking back at one piece of evidence that led us to the Law of Combining Volumes, we found that 1 L of hydrogen plus 1 L of chlorine yields 2 L of hydrogen chloride. Using the conclusion of Avogadro's Law, the volume ratio and the particle ratio must be the same. This seems to say that 1 hydrogen atom plus 1 chlorine atom makes 2 hydrogen chloride molecules. But this can't be! How could we make 2 identical molecules of hydrogen chloride from a single chlorine atom and a single hydrogen atom? This would require us to divide each hydrogen and chlorine atom, violating the postulates of the Atomic Molecular Theory. There is one solution to this problem, as was recognized by Avogadro. We have to be able to divide a hydrogen gas particle into two identical pieces. This means that a hydrogen gas particle must contain an even number of hydrogen atoms, most simply, two. This says that hydrogen gas exists as hydrogen molecules, and each hydrogen molecule contains two hydrogen atoms. The same conclusions apply to chlorine: a chlorine gas molecule must contain two chlorine atoms. If these conclusions are correct, then one hydrogen molecule, H2 , can react with one chlorine molecule, Cl2 , to form two hydrogen chloride molecules, HCl. The ratio of the reactant particles and the product particles is then the same as the ratio of the reactant gas volumes and the product gas volumes. This is a wonderful result because we have now determined the molecular formula of hydrogen chloride, HCl. We have found a way to count the numbers of atoms in the reaction, at least in ratio, by measuring the volumes of the gases that react and that are produced. This gives us a chemical reaction showing the atoms and molecules that participate in the reaction in the correct ratio: 1 H2 molecule + 1 Cl2 molecule → 2 HCl molecules This chemical equation is consistent with all of our known observations and the postulates of the Atomic Molecular Theory. Available for free at Connexions

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CHAPTER 2. ATOMIC MASSES AND MOLECULAR FORMULAS

Since the Law of Combining Volumes is a general result, we can look at many chemical reactions with the same analysis. Let's apply this to the hydrogen and oxygen reaction discussed earlier. Remember that 2 L of hydrogen react with 1 L of oxygen to form 2 L of water vapor. This means that two particles of hydrogen (which we know to be H2 ) react with one particle of oxygen to form two particles of water. Once again, we have the problem that one atom of oxygen cannot make two molecules of water. Therefore, an oxygen gas particle cannot be an oxygen atom, so oxygen gas exists as oxygen molecules, O2 . Since two H2 molecules react with one O2 molecule to form two water molecules, each water molecule must be H2 O. We can write the chemical equation: 2 H2 molecules + 1 O2 molecule → 2 H2 O molecules We can use these observations to nally solve the riddle which is posed in Table 2.2: Possible Molecular Formulae for Nitrogen Oxides. We need to observe the volumes of oxygen and nitrogen which react to form Oxides A, B, and C. In separate experiments, we nd: 1 L N2 + 2 L O2 → 2 L Oxide A 1 L N2 + 1 L O2 → 2 L Oxide B 2 L N2 + 1 L O2 → 2 L Oxide C (At this point, it is pretty clear from the data and using our previous reasoning that nitrogen gas must consist of nitrogen molecules, N2 , since 1 L of nitrogen gas can make 2 L of Oxide B.) From these data, we can conclude that Oxide B has molecular formula NO, since 1 L of oxygen plus 1 L of nitrogen produces 2 L of Oxide B with nothing left over. Similarly and with the use of Table 2.2: Possible Molecular Formulae for Nitrogen Oxides, we can say that Oxide A is NO2 and Oxide C is N2 O.

2.5 Observation 2: Relative Atomic Masses In the Introduction, we presented a dilemma in developing the Atomic Molecular Theory. To nd the molecular formula of a compound, we needed to nd the relative atomic masses. And to nd the relative atomic masses, we needed to nd the molecular formula of a compound. Using Avogadro's Law, we have found a way to break out of this dilemma. By measuring gas volumes during reactions, we can essentially count the numbers of atoms in a molecule, giving us the molecular formula. Our task now is to use this information to nd atomic masses. We can begin by looking at the data in Table 2.1: Mass Relationships of Simple Compounds of Nitrogen and Oxygen and focusing on Oxide B at rst. We know now that Oxide B has molecular formula NO. As such, it is given the name Nitric Monoxide, or more commonly Nitric Oxide. We also know from Table 2.1: Mass Relationships of Simple Compounds of Nitrogen and Oxygen that the mass ratio of oxygen to nitrogen in NO is 1.14 to 1.00. Since there are equal numbers of nitrogen atoms and oxygen atoms in any sample of NO, then the mass ratio of an oxygen atom to a nitrogen atom is also 1.14 to 1.00. Stated dierently, an oxygen atom has mass 1.14 times greater than a nitrogen atom. This is a good start, but now we need more elements. To bring in hydrogen, we can analyze the data from Table 2.2: Possible Molecular Formulae for Nitrogen Oxides in the rst Concept Development Study which gives the mass ratio of oxygen and hydrogen in water. That data shows that the mass ratio of oxygen to hydrogen is 7.93 to 1.00. But we found in the previous section that the molecular formula of water is H2 O. This means that in a sample of water there are twice as many hydrogen atoms as there are oxygen atoms. Therefore, the ratio of the mass of one oxygen atom to one hydrogen atom must be 7.93 to 0.50, or 15.86 to 1.00. These atomic mass ratios need to be consistent with each other, since the masses of the atoms of an element are always the same. So if the ratio of one hydrogen to one oxygen is 1.00 to 15.86, and the ratio of one nitrogen to one oxygen is 1.00 to 1.14, then the ratio of one hydrogen to one nitrogen must be 1.00 to 13.91. We should be able to check this by looking at the hydrogen-nitrogen compound ammonia, also listed in Table 2.2: Possible Molecular Formulae for Nitrogen Oxides of the previous Concept Development Study. There we nd that the mass ratio of nitrogen to hydrogen is 4.65 to 1.00. Clearly, ammonia is not NH. To nd the molecular formula of ammonia, we need data from the Law of Combining Volumes. Experimental data reveal that 1 L of N2 reacts with 3 L of H2 to produce 2 L of ammonia. From this, we should be able Available for free at Connexions

15 to conclude that an ammonia molecule has the molecular formula NH3 . Therefore, in a sample of ammonia, there are three times as many hydrogen atoms as there are nitrogen atoms. This means that the ratio of the mass of a nitrogen atom to a hydrogen atom is 3*4.65 to 1.00, or 13.95 to 1.00. We now have enough data to say that hydrogen, nitrogen, and oxygen atoms have mass ratio of 1.00:13.95:15.86.

2.6 Observation 3: Atomic Masses for Non-Gaseous Elements The next element we would certainly like to have an atomic mass for would be carbon, and we would certainly like to be able to determine molecular formulas for carbon containing compounds. We have data from Table 2.3: Mass Relationships of Simple Compounds of Carbon and Oxygen in the previous CDS on compounds of hydrogen and carbon. But our analysis is not going to work this time. The Law of Combining Volumes and Avogadro's Law in combination allow us to count atoms and nd molecular formulas, but only for elements and compounds which are gases. Carbon is not a gas. It exists in several dierent elemental forms, but all are solid at normal temperatures and even at very high temperatures. This means that we need to work harder and add some additional observations to our work. Let's start with the two most common oxides of carbon, which for now we will give the names Oxide A and Oxide B. (Their real names, carbon monoxide and carbon dioxide, are based on assuming that we already know their molecular formulas. But we don't know these, so we'll stick with these code names for now.) Here are the data for the mass relationships from the Law of Multiple Proportions:

Mass Relationships of Simple Compounds of Carbon and Oxygen Compound Total Mass (g) Mass of Carbon (g) Mass of Oxygen (g) Oxide A 2.33 1.00 1.33 Oxide B 3.66 1.00 2.66 Table 2.3

It is clear to see the Law of Multiple Proportions here since Oxide B has exactly twice the mass of Oxide A for the same mass of carbon. Now we would like to observe the Law of Combining Volumes and apply Avogadro's Law, but carbon is not a gas. But we can at least look at the ratios of the volume of oxygen and the volume of the oxides produced. For Oxide A, 1 L of oxygen will produce 2 L of Oxide A. For Oxide B, 1 L of oxygen will produce 1 L of Oxide B. From these data, we can see that an Oxide A molecule contains one oxygen atom, since a single O2 molecule makes two Oxide A molecules. We can also see that an Oxide B molecule contains two oxygen atoms, since a single O2 molecule produces a single Oxide B molecule. Now we know part of the molecular formulas each of the oxides, but we don't know the number of carbon atoms in each. Oxide A could be CO, C2 O, C3 O, etc., and Oxide B could be CO2 , C2 O2 , etc. Our only way to proceed is with Avogadro's Law. We can use this to determine the relative mass of each oxide molecule, even if we can't determine the relative mass of the carbon atoms. The mass of 1 L of Oxide A is less than the mass of 1 L of oxygen. This is because, even though equal volumes contain the same number of particles, the particles themselves have dierent masses. In fact, since 1 L of each gas contains the same number of particles, then the mass ratio of 1 L of Oxide A to oxygen is exactly the same as the mass ratio of one Oxide A molecule to one oxygen molecule. An experimental measurement shows that the mass of 1 L of Oxide A is 87.5% of the mass of 1 L of oxygen. So the mass of one particle of Oxide A must be 87.5% of the mass of an O2 molecule. On the relative scale found in the previous section, the mass of one oxygen molecule is 15.86*2 = 31.72. So a particle of Oxide A must have relative mass 0.875*31.72 = 27.76. Remember though that an Oxide A contains just one oxygen atom. So, of this relative mass, 15.86 belongs to the oxygen atom, leaving 11.90 for the carbon. Is this the relative mass of a carbon atom? We don't know because we don't know whether a molecule of Oxide A contains 1, 2, or 3 atoms of carbon (or for that matter, any number). What we can say is that Available for free at Connexions

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CHAPTER 2. ATOMIC MASSES AND MOLECULAR FORMULAS

the mass of a carbon atom is either 11.9, one-half of that, one-third of that, or some integer fraction of that depending on how many carbon atoms are in Oxide A. How can we determine the right number? The answer is found from persistence. We can repeat this same measurement and calculation for Oxide B. The mass of 1 L of Oxide B is 1.375 times greater than the mass of 1 L of oxygen. Therefore, one particle of Oxide B has relative mass 43.59. But Oxide B molecules contain two oxygen atoms, so the part of this mass which is carbon is 43.59 minus the relative mass of the two oxygen atoms. This gives the mass of carbon in Oxide B as 11.9. This is the same mass of carbon that we found in Oxide A. Therefore, Oxide A and Oxide B have exactly the same number of carbon atoms. Perhaps, the number of carbon atoms in each is just one, but we can't be sure from this data. To convince ourselves, we should repeat this process for many gaseous compounds of carbon, including ones that contain hydrogen and oxygen both. In every experiment, we nd that the relative mass of carbon in each molecule is either 11.9 or a simple multiple of 11.9. It is never less than 11.9. We can conclude that the relative mass of a carbon atom on the same scale we have been using is 11.9. We can also conclude that Oxide A and Oxide B each contain one carbon atom, so the molecular formulas of Oxide A and Oxide B are CO and CO2 . This procedure can be used to nd molecular formulas of compounds containing other non-gaseous elements. This is the actual procedure that was used around 1850 to provide the rst set of relative atomic masses and the rst denite molecular formulas for common compounds. This is a major stride forward from the postulates of the Atomic Molecular Theory. As one last step, we note that the standard agreed on by chemists for the relative atomic masses does not take hydrogen to have mass 1.00. Rather, on the agreed upon scale, the relative atomic mass of hydrogen is 1.008, that of carbon is 12.01, that of nitrogen is 14.01, and that of oxygen is 15.999. These ratios are the same as the ones we observed in our calculations.

2.7 Chemical Algebra: Stoichiometry In the Introduction (Section 2.1: Introduction), we decided that we could determine molecular formulas if we knew the relative atomic masses. At that point, we didn't have the relative atomic masses, but now we do. Once we know all the relative atomic masses, we no longer need the Law of Combining Volumes and Avogadro's Law to determine molecular formulas. Let's show this by an example, with a compound which contains only carbon, oxygen, and hydrogen. An analysis from the Law of Denite Proportions gives us that the compound is 40.0% carbon, 53.3% oxygen, and 6.7% hydrogen by mass. In other words, if we have a 100.0 g sample of the compound, it consists of 40.0 g of carbon, 53.3 g of oxygen and 6.7 g of hydrogen. But we also know that the relative masses of carbon, oxygen, and hydrogen are 12.01:15.99:1.008. This will allow us to determine the relative numbers of atoms of each type in the compound. To do this, we create a method of chemical algebra. Let's start by assuming that we have exactly N atoms of carbon, N atoms of hydrogen, and N atoms of oxygen. N is some very large number, and it doesn't matter what it is, as long as we have taken the same N for all three elements. The relative mass of 1 carbon atom to 1 hydrogen atom is 12.01 to 1.008. Therefore the relative mass of N carbon atoms to N hydrogen atoms is also 12.01 to 1.008. Let's pick a very specic N: let's make N be whatever number it is such that that a sample of N carbon atoms has mass 12.01g. Interestingly, we don't need to know what N is  we just need to nd a sample of carbon which has a mass of 12.01 g. What is the mass of N hydrogen atoms (for the exact same N)? It must be 1.008 g, since each hydrogen atom has mass ratio to each carbon atom 1.008 to 12.01. Therefore, if we weigh out a sample of carbon with mass 12.01 g and a sample of hydrogen with mass 1.008 g, we know that we have exactly the same number of atoms of each type. Since this seems like a useful number of atoms, we will give it a name. N is called a mole of atoms. We don't need to know what N is to know that we can nd a mole of atoms simply by nding the mass of a sample: 12.01 g of carbon, 1.008 g of hydrogen, 15.99 g of oxygen, and so on. (For historical reasons, the value of N which is a mole of atoms is called Avogadro's number, in his honor but not because he discovered Available for free at Connexions

17 the number. Avogadro's number is given the symbol NA . The number of particles in a mole is approximately 6.022×1023 , although we will almost never need this number when doing chemical calculations.) Since we know the mass of one mole of a substance, we can nd the number of moles in a sample of that substance just by nding the mass. Consider a sample of carbon with mass 24.02 g. This is twice the mass of one mole, so it must contain twice the number of particles as one mole. This must be two moles of particles. That example was easy, but what if we have 30.02 g of carbon? Since one mole has mass 12.01 g, then 30.02 g must contain 30.02/12.01 moles = 2.5 moles. Even more generally, then, if we have a sample of an element has mass m and the atomic mass of the element is M, the number of moles of atoms, n, is m n= M

Since one mole contains a xed number of particles, regardless of the type of particle, calculating the number of moles n is a way of counting the number of particles in a sample with mass m. For example, in the 100.0 g sample of the compound above, we have 40.0 g of carbon, 53.3 g of oxygen, and 6.7 g of hydrogen. We can calculate the number of moles of atoms of each element using the equation above: 0g nC = 12.40. 0g/mol =3.33moles 3g nO = 16.53. 0g/mol =3.33moles 6.7g nH = 1.0g/ =6.67moles mol A mole is a xed number of particles. Therefore, the ratio of the numbers of moles is also the same as the ratio of the numbers of atoms. In the data above, this means that the ratio of the number of moles of carbon, oxygen, and hydrogen is 1:1:2, and therefore the ratio of the three types of atoms in the compound is also 1:1:2. This suggests that the compound has molecular formula COH2 . However, this is just the ratio of the atoms of each type, and does not give the number of atoms of each type. Thus the molecular formula could just as easily be C2 O2 H4 or C3 O3 H6 . Since the formula COH2 is based on empirical mass ratio data, we refer to this as the empirical formula of the compound. To determine the molecular formula, we need to determine the relative mass of a molecule of the compound, i.e. the molecular mass. One way to do so is based on the Law of Combining Volumes, Avogadro's Hypothesis, and the Ideal Gas Law. To illustrate, however, if we were to nd that the relative mass of one molecule of the compound is 60.0, we could conclude that the molecular formula is C2 O2 H4 . Counting the relative number of particles in a sample of a substance by measuring the mass and calculating the number of moles allows us to do chemical algebra, calculations of the masses of materials that react and are produced during chemical reactions. This is easiest to see with an example. Some of the most common chemical reactions are those in which compounds of hydrogen and carbon, called hydrocarbons, are burned in oxygen gas to form carbon dioxide and water. The simplest hydrocarbon is methane, and using the methods of this study, we can nd that methane has the molecular formula, CH4 . The chemical equation which represents the burning of methane is: 1 CH4 molecule + 2 O2 molecules → 1 CO2 molecule + 2 H2 O molecules It is important to note that the number of atoms of each type is conserved during the chemical reaction. The reactants and products both contain 1 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms. This is called a balanced chemical equation, and it expresses the postulate of the Atomic Molecular Theory that the numbers of atoms of each element does not change during a chemical reaction. In chemical algebra, we can ask and answer questions such as, If we burn 1.0 kg of methane, what is the mass of carbon dioxide which is produced? Such a question would clearly be of importance in understanding the production of greenhouse gases like CO2 . The chemical equation above expresses a relationship between the number of molecules of methane which are burned and the number of molecules of CO2 produced. From the equation, each molecule of CH4 produces one molecule of CO2 . Therefore, if we knew how many molecules of CH4 we have in a sample, we know how many molecules of CO2 we will produce. The chemical equation works for any number of molecules. If we burn N molecules of CH4 , we produce N molecules of CO2 . This will work no matter what N is. Therefore, we can say that 1 mole of CH4 molecules will produce 1 mole of CO2 molecules. The chemical equation works just as well for moles as it does for molecules, since 1 mole is just a xed number of molecules. And we know how to calculate the number of moles from a measurement of the mass of the sample. Available for free at Connexions

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CHAPTER 2. ATOMIC MASSES AND MOLECULAR FORMULAS

Recall that we are interested in what happens when we burn 1.00 kg = 1000 g of methane. We just need to know the mass of a mole of methane. Since one molecule of methane has relative mass 16.0, then one mole of methane has mass 16.0 g/mol. Then the number of moles in 1000 g of methane can be calculated by dividing by the mass of 1.0 mole of methane: g nCH = 16.1000 0g/mol =62.5moles This means we have counted the number of particles of CH4 in our sample. And we know that the number of particles of CO2 produced must be the same as this, because the chemical equation shows us the 1:1 ratio of CH4 to CO2 . So, 62.5 moles of CO2 are produced by this reaction and this 1.0 kg sample. We are usually more interested in the mass of the product, and we can calculate this, too. The mass of one mole of CO2 is found from the mass of one mole of C and two moles of O, and is therefore 44.0 g. This is the mass for one mole. The mass for 62.5 moles will be mCO =nCO MCO (62.5moles) (44.0g/mol) =2750g=2.75kg Therefore, for every 1 kg of methane burned, we produce 2.75 kg of CO2 . The important conclusion from this example of chemical algebra is that it is possible to calculate masses of products from masses of reactants. We do so by using a balanced chemical equation and by understanding that the equation gives us the ratio of moles of reacting materials just as it gives us the ratio of molecules of reacting materials. This is because numbers of moles and numbers of molecules are simply dierent ways of counting the number of particles. Chemical algebra is usually referred to as stoichiometry, a somewhat intimidating term that makes the calculations seem harder and more abstract than they are. We really only need to remember two things. First, from the Atomic Molecular Theory, a chemical reaction can be represented by a balanced chemical equation which conserves the numbers of atoms of each element. Second, the balanced equation provides the ratio of the number of product molecules to the number of reactant molecules, either in numbers of molecules or numbers of moles. Thus, we can solve problems eciently by calculating the number of moles. A nal interesting note about Avogadro's number is helpful in understanding what 1 mole is. A question often asked is, where did the number 6.022×1023 come from? If we wanted to pick a very large number for the number of particles in a mole, why didn't we pick something easier to remember, like 6×1023 , or even 1×1023 ? The value of Avogadro's number comes from the fact that we chose 1 mole to be the number of carbon atoms in 12.01g of carbon. Since 1 carbon atom has mass 12.01 amu, then the mass of NA carbon atoms is NA ×12.01 amu. But 1 mole of carbon atoms has mass 12.01 g, so 12.01 g must equal NA ×12.01 amu: 12.01 g = NA ×12.01 amu This means that 1 g = NA amu This shows that Avogadro's number is just the conversion factor for mass between grams and amu. We didn't randomly pick Avogadro's number. Rather, we picked the unit of mass amu, and it turns out that there are Avogadro's number of amu in one gram. 4

2

2

2

2.8 Review and Discussion Questions 1. State the Law of Combining Volumes and provide an example of your own construction which demonstrates this law. 2. Explain how the Law of Combining Volumes, combined with the Atomic Molecular Theory, leads directly to Avogadro's hypothesis that equal volumes of gas at equal temperatures and pressure contain equal numbers of particles. 3. Use Avogadro's hypothesis to demonstrate that oxygen gas molecules cannot be monatomic. 4. The density of water vapor at room temperature and atmospheric pressure is 0.737 g/L. Compound A is 80.0% carbon by mass, and 20.0% hydrogen. Compound B is 83.3% carbon by mass and 16.7% hydrogen. The density of gaseous Compound A is 1.227 g/L, and the density of Compound B is 2.948 g/L. Show how these data can be used to determine the molar masses of Compounds A and B, assuming that water has molecular mass 18. Available for free at Connexions

19 5. From the results in Problem 4, determine the mass of carbon in a molecule of Compound A and in a molecule of Compound B. Explain how these results indicate that a carbon atom has atomic mass 12. 6. Explain the utility of calculating the number of moles in a sample of a substance. 7. Explain how we can conclude that 28g of nitrogen gas (N2 ) contains exactly as many molecules as 32g of oxygen gas (O2 ), even though we cannot possibly count this number. By John S. Hutchinson, Rice University, 2011

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CHAPTER 2. ATOMIC MASSES AND MOLECULAR FORMULAS

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Chapter 3

Structure of an Atom

1

3.1 Introduction The Atomic Molecular Theory is very powerful in helping us understand matter. We can easily understand the dierences between the elements and all other compounds by thinking about the particles which make them up. Elements contain all identical atoms, and compounds contain identical molecules, each composed of denite small numbers of atoms of the elements that make up the compound. We can also do chemical algebra calculations, which allow us to make predictions about the masses of materials that are involved in chemical reactions. Without further observations, though, this is about as far as we can go because our knowledge about the properties of atoms is very limited. We only know that atoms of dierent elements have dierent masses, and we know the relative atomic masses of the elements. But this does not give us any insight into why the atoms of dierent elements have such diering chemical properties. For example, carbon and nitrogen have very similar atomic masses. However, carbon in one of its forms, diamond, is a very hard, unreactive solid and in another of its solid forms, coal, carbon will burn in oxygen. Nitrogen, though, is a gas and is fairly unreactive with oxygen except at very high temperatures. Clearly, carbon atoms interact with each other very dierently than nitrogen atoms interact with each other. Without knowing anything else about atoms, we might imagine them as tiny hard spheres. But this idea will not help us understand their chemistry, such as why they form the compounds that they do. What does it mean for hydrogen and carbon to form the compound methane? We know that methane's molecular formula is CH4 , but how do these ve atoms hold together in a molecule? It seems reasonable to imagine that the attractions are due to magnetic forces or electrical forces between the atoms. However, atoms of hydrogen and carbon must be electrically neutral, since bulk samples of hydrogen and carbon are electrically neutral. They are also non-magnetic. How do neutral, non-magnetic atoms interact with each other? Another riddle is why they form in the particular ratios that they do. CH4 is a compound but there isn't a compound with molecular formula CH5 . Understanding these diering elemental and molecular properties requires us to have a deeper understanding of the properties of individual atoms. Since we cannot understand these properties by thinking of the atoms as individual hard spheres, we need to investigate the structure of the atom.

3.2 Foundation In this study, we will assume that we know the postulates of the Atomic Molecular Theory and our measurements of relative atomic masses. We know that an element is composed of individual atoms with identical masses, and we know that the atoms of dierent elements have dierent masses, which have been measured. 1 This content is available online at .

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CHAPTER 3. STRUCTURE OF AN ATOM

We will also assume our understanding from a previous study that electricity consists of individual charged particles called electrons, which are assigned a negative charge. The mass of the electron is quite small relative to the mass of an atom. Atoms contain electrons, and electrons can be added to or removed from atoms by applying an electrical current. We noted in the introduction that atoms in elemental form are neutral, or without charge. Since atoms contain negatively charged electrons, each atom must also contain a positive charge which exactly equals the total charge on the electrons in the atom. Since electrons are particles, there has to be an integer number of negative charges, so there also has to be an integer number of positive charges in each atom. These statements are important to keep in mind, but they don't help us very much without more data. We don't know how many electrons are in each atom, and therefore we don't know how many positive charges are in each atom. We do not even know whether that number of electrons is the same or dierent for dierent atoms of dierent elements. In addition, we don't know how these positive and negative charges are arranged in each atom. They might be clustered together in some sort of ball, they might be paired o together, they might be randomly arranged, or there may be some other arrangement that we don't expect. In this concept development study, we will determine the arrangement of the charged particles in an atom, and we will determine the numbers of charged particles in the atoms of each element.

3.3 Observation 1: Scattering of α particles by atoms To nd out what the inside of an atom looks like, we perform a scattering experiment. This involves shooting charged particles at atoms and watching what happens. Depending on how the charged particles are arranged inside the atom, these charges should scatter the particles we shoot. If we look at the pattern of the scattering, we should be able to infer what the arrangement of the charges inside the atom looks like. It would seem very hard to shoot particles at individual atoms, though, since these are very small, impossible to see, and possibly even elusive. One way to do this is to take a very thin sheet of a metal, so thin that the thickness of the metal will not be very many atoms across. Gold is a good choice of metal, since it is easy to hammer out to a very thin sheet and it is very unreactive, so it is possible to make it quite pure. In this experiment, the thickness of the gold foil will be only about 10-4 cm, sometimes called 1 micron. This is less than one-twentieth of the thickness of one human hair, so this is very thin indeed. We need particles to shoot at this thin foil, preferably charged particles which will interact with the positive and negative charges in the gold atoms. A good choice is the α particle, which is positively charged and much more massive than an electron. In the experiment, we will re a beam of α particles directly at the gold foil, and then we will observe where the α particles go after interacting with the gold atoms. They might pass through the foil, they might be deected somewhat as they pass through, or they might even rebound in various angles back to the source of the beam.

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Figure 3.1: Diagram of Rutherford's gold foil experiment showing the results of shooting

A

particles

at a very thin sheet of gold foil.

The experimental data for the scattering of the α particles shows three primary results, as depicted in Figure 3.1. First, and perhaps most surprisingly, the greatest number of the α particles pass directly through the gold foil without any deection in their paths. Second, a much smaller number of the α particles experience small deections in their paths. And third, a very small fraction of the α particles (perhaps 1 in 50,000 or more) are deected back in the direction of the beam they came from. For now, we'll focus on the rst and third observations. It is surprising that the largest number of the α particles pass through the gold foil as if it is not there. In fact, it is just as if the gold foil consisted mostly of empty space, and most of the α particles seem to pass through that empty space. This seems strange, since the gold foil is certainly solid and doesn't appear to be empty space. The third result gives a striking contrast to this result. A tiny fraction of the α particles must encounter something other than empty space. To rebound, an α particle must hit something much more massive than itself. We might think that this is a gold atom, which is much heavier than an α particle, but to be consistent with the rst observation, most of the gold atom must be empty space. Thinking about these two observations leads us to a simple model for the atom. Each gold atom must be mostly empty space and most of the mass of the gold atom must be concentrated into a very small fraction of the volume of the atom. We will call this concentrated mass the nucleus. A careful calculation based on the fraction of α particles which pass through and the fraction which rebound tells us that the diameter of the nucleus is about 100,000 times smaller than the diameter of the atom itself. This is an amazing result! Although an atom is very small, the nucleus is very much smaller than that, as analogously depicted in the diagram in Figure 3.2.

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CHAPTER 3. STRUCTURE OF AN ATOM

Figure 3.2: This map illustrates the size of an atom if the nucleus were a standard-sized basketball at

Lovett Hall on the Rice University campus. The atom is 100,000 times large than the nucleus, so on this map this model of an atom has a radius of 11.9 km.

This is an interesting model that seems to account for two of our three observations, but it is a puzzling model too. If the mass of the atom is concentrated in such a small space, what occupies the rest of the volume of the atom? The second observation from the experiment helps us understand this. Since a small number of the positively charged α particles are deected in their paths, this suggests that they come close to but do not run into something positive in the atom. Since these α particles are deected, whatever they come close to must be more massive than they are. This means that they are coming close to the nucleus and the nucleus must be positively charged. This gives us a clue about what occupies the vast empty space of the atom. If the positive charges of the atom are concentrated in the nucleus, the negative charges in the atom, the electrons, must be in the much larger space of the atom outside of the nucleus. This nuclear model of the atom then accounts for all three observations in the α particles gold foil scattering experiment. Most of the volume of the atom is empty space in which negatively charged electrons move about. Most of the mass and all of the positive charge of the atom is concentrated in a nucleus, which is tiny in comparison to the atom.

3.4 Observation 2: X-ray emissions from atoms It is interesting to know that each atom has a nuclear structure, particularly since we now know that the positive and negative charges in the atom are separated into dierent parts of the atom. The interaction of these positive and negative charges must be what determines the properties of each type of atom, including the types of molecules that they tend to form. Even though each atom is neutral, it might be possible for the positive charges on one atom to interact with the negative charges on another atom, and vice versa. To show this, we would need to know how many positive and negative charges there are in each atom. It seems reasonable to assume that these numbers are dierent for atoms of dierent elements, but we don't know that without making more observations. At this point, we know that atoms contain electrons, and that there are an integer number of these. We also know that each atom contains an equal number of positive charges, all of which are in the nucleus. It makes sense that, since the positive charge is an integer, there must be particles of positive charge in the nucleus, and we will now call these particles protons. This doesn't make our life any easier, but it does clarify our language. Our problem now boils down to nding out how many protons and how many electrons there are in each atom of each type. These integers are the same number for each atom, of course, since each atom is neutral. Available for free at Connexions

25 Finding this integer for each element seems quite dicult, and the experiment which reveals the number to us is a strange one which doesn't seem related to the question at all. When materials are placed in an electrical discharge, they commonly emit light or electromagnetic radiation. Not all of this light is visible. Some of this light is high-energy light called ultraviolet radiation, and some of it is even higher energy radiation called x-rays. We can tell the dierence between dierent types of radiation by the frequency of the radiation, a property that tells us how rapidly the electromagnetic eld of the radiation oscillates. Dierent frequencies of radiation can be separated by passing the radiation through dierent materials. This is how a prism works, for example. We can also separate the dierent frequencies with something called a diraction grating, which can be either a solid with parallel grooves or a transparent solid with closely spaced lines. As common examples of simple diraction gratings, compact discs and DVDs have grooves in parallel to each other and can be used to produce a visible rainbow. A white light source will reveal the dierent colors of light, each with its own frequency, when the light is reected o of the surface of the disc. This separation of light into dierent frequencies is very useful in a type of experiment called spectroscopy. There are many types of spectroscopy, but in most cases, a sample of a material is energized in some way, often in an electrical discharge, and the sample emits light. This is an everyday observation. For example, heated objects tend to glow, which means that they are emitting light. In spectroscopy, the light which the sample emits is separated by a prism or a grating and a wonderful result occurs. Each dierent element or compound has its own characteristic set of frequencies of light which are emitted. In other words, only certain frequencies of light are emitted by each substance, and the set of frequencies for each substance is unique to that substance. Spectroscopy has great value in science, because we can identify that a sample contains a particular compound or element just by looking at the frequencies of light that the sample emits. This is how we know the composition of distant stars, for example. We need to focus on atoms, so we'll look only at the spectroscopy of pure elements. Each element has a characteristic spectrum of frequencies. We will focus on the light frequencies which are x-rays and, even more carefully, for each element we will focus on the lowest frequency x-ray which each atom emits. These are given for many of the elements in Table 3.1: X-ray Frequencies of Atoms. The ordering of the elements in Table 3.1: X-ray Frequencies of Atoms is important. We could have listed them alphabetically by name. Instead, we have listed them in their order of increasing relative mass. In the set of elements given here, lithium is the least massive element so it is at the top, then beryllium, and so forth. Note that the x-ray frequency emitted by each atom increases as the masses of the atoms increase moving down Table 3.1: X-ray Frequencies of Atoms. (There is one exception to our order in Table 3.1: X-ray Frequencies of Atoms: Argon is slightly more massive than Potassium, but we have put Argon before Potassium for reasons that we will discuss later.)

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CHAPTER 3. STRUCTURE OF AN ATOM

X-ray Frequencies of Atoms Atomic Number Element Name X-ray frequency (10 16 s-1) 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminum Silicon Phosphorous Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium

1.3158 2.6316 4.4379 6.7114 9.4937 12.701 16.376 20.534 25.189 30.334 35.971 42.008 48.701 55.814 63.416 71.518 80.118 89.242 98.873 109.01 119.65 130.80 142.47 154.65 167.33 180.53 194.25 208.48 223.21 238.47

Table 3.1

We have also added something interesting in Table 3.1: X-ray Frequencies of Atoms, the atomic number. This number is just the ranking of the elements in order of increasing mass. The atomic number then is just another name we have given to each element with no more signicance than the names we have given. (Note Available for free at Connexions

27 again that Argon is placed before Potassium. There are other reasons for this, but this also makes sense in the data, since this way, the frequencies of the x-rays increase.) Other than the fact that the x-ray frequencies consistently increase, there doesn't seem to be any other obvious pattern in the data. We could look for a pattern by plotting the x-ray frequency versus the masses of the atoms, but this doesn't show anything additional. Surprisingly, though, if we plot the x-ray frequency versus the atomic number, a clear pattern emerges, as we see in Figure 3.3.

Figure 3.3: X-ray frequency vs. atomic number for the elements lithium to germanium.

The frequencies don't just increase with atomic number, they increase with a very smooth function. This type of graph might look familiar to you. It looks similar to a graph of y = x2 , called a parabola. Is the graph in Figure 3.1 a parabola? To nd out, we need to know whether the x-ray frequency is a function of the atomic number squared. One way to do this is to plot the square root of the x-ray frequency versus the atomic number. This would be like plotting the square root of y versus x to see if y = x2 . This new graph is shown in Figure 3.4.

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CHAPTER 3. STRUCTURE OF AN ATOM

Figure 3.4:

The square root of x-ray frequencies vs.

atomic number for the elements lithium to

germanium.

The graph in Figure 3.2 is very signicant. It shows that there is a simple relationship between the x-ray frequency emitted by each atom and the atomic number of the atom. But this is a strange result. The x-ray frequency is a physical property of each atom. The atomic number is just an integer that we assigned to each element based on its mass ranking. This is a surprise! Why would a physical property match up so strongly with a non-physical property like an integer in the mass rankings? There is only one possibility: the atomic number must also be a physical property of the atom. It is not an arbitrary number, even though we assigned it that way initially. It is a property of the atom. The atomic number is also a special property of the atom because it is an integer, and integers are used to count objects or particles. This means that the atomic number must be counting some integer number of objects or particles in each atom, and that number must be unique for each atom. But we know that there is an integer that describes each atom: the number of protons and electrons that each atom has. Our conclusion is therefore that the atomic number is a property of each atom which is equal to the number of protons in each atom and is also equal to the number of electrons in each atom. We have found a way to count the number of charges in each atom! There are two remarkable results to mention here. The rst is that each atom has its own characteristic number of protons and electrons. This is not shared with any other atom. This suggests that these numbers must be important in determining the chemical and physical properties of the atom. The second is that every integer number of protons and electrons is accounted for. There are no gaps or breaks in Table 3.1: X-ray Frequencies of Atoms or in Figure 3.1 or Figure 3.2. We have them all. There are no missing elements in the range from 3 to 32, nor for that matter from 1 to 114. This is why there are only a small number of elements (about 90 naturally occurring ones). We now know what is unique about the atoms of dierent elements. It does make sense that atoms with dierent numbers of charged particles would interact dierently, forming compounds with dierent chemical and physical properties. But we might imagine that atoms with similar (but not equal) numbers of charged particles would have fairly similar chemical and physical properties. As we will explore in much more detail later, this is not at all the case. In fact, it is possible to compare two elements with very similar atomic numbers, say Neon and Sodium, and nd that their chemical properties are completely dierent. Neon is a gas, even at extremely low temperatures, and is so unreactive that it does not form compounds with any other elements. Sodium is a solid which is so reactive that its reactions with other elements are often violent and energetic. It seems clear that just knowing how many electrons an atom has is not sucient to make Available for free at Connexions

29 predictions about how that atom might behave. We will need to rene our model with further observations.

3.5 Review and Discussion Questions 1. Explain how the scattering of α particles from gold foil reveals that an atom contains a massive, positively charged nucleus whose size is much smaller than that of an atom. 2. Explain the signicance of the relationship between the frequency of x-ray emission from each atom and the atomic ranking of that atom in the periodic table. By John S. Hutchinson, Rice University, 2011

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CHAPTER 3. STRUCTURE OF AN ATOM

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Chapter 4

Electron Shell Model of an Atom

1

4.1 Introduction What more could we want to know about the structure of an atom? We know that atoms contain positively and negatively charged particles, and that the number of these charges in each atom is dierent for each element. We also know that the positive charges are concentrated in a tiny nucleus, and that the electrons move around the nucleus in a space that is much, much larger than the nucleus. However, some of the most important questions we asked in the previous Concept Development Study are still unanswered. Remember that we saw that carbon and nitrogen have very similar atomic masses. Now we can add that these elements have very similar atomic numbers, so their atoms have similar numbers of protons and electrons. But carbon and nitrogen are, in most chemical and physical ways, very dierent. Similarly, some elements like sodium and potassium have very dierent atomic numbers but have quite similar chemical and physical properties. It seems that comparing the properties of two dierent atoms is not very easy to understand just from comparing the numbers of protons and electrons the atoms contain. To continue to understand the answers to these questions, we need even more detail about the structure of each type of atom.

4.2 Foundation In this study, we will assume that we know the postulates of the Atomic Molecular Theory and our measurements of relative atomic masses. We know that an element is composed of individual atoms with identical masses, and we know that the atoms of dierent elements have dierent masses, which have been measured. We will also assume that we know that structure of an atom, with a tiny, massive, positively charged nucleus surrounded by a much larger empty space in which electrons move. The positive charge on the nucleus is equal to the number of protons in the nucleus and, in a neutral atom, is also equal to the number of electrons moving about the nucleus. The number, called the atomic number, is unique for each type of atom. No two elements have the same atomic number, and amongst the naturally occurring elements, no atomic number is skipped: for every integer up to 118 we know an element with that atomic number. In this study, we will need a very important observation borrowed from the study of Physics. We will use Coulomb's Law to describe the interaction of charged particles. Coulomb's Law is an algebraic expression which relates the strength of the interaction between two charged particles to the sizes of the charges on the particles and the distance between them. We can think of the strength of the interaction between particles as either the force that one particle exerts on the other particle or the potential energy which exists when the two particles interact with each other. We will focus on the potential energy, which we will call V. Let's think of two particles, one with charge q1 and the other with charge q2 . These charges can be either positive 1 This content is available online at .

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CHAPTER 4. ELECTRON SHELL MODEL OF AN ATOM

or negative, depending on the properties of the particles. Let's place the two particles a distance r away from each other. Then the potential energy of interaction between these two charged particles is: 2) V = (q1 )(q r

This is Coulomb's Law. We will very rarely do any calculations with this equation. Instead, we will apply it to understand when V is expected to be a large number or a small number, positive or negative. When V is a large negative number, the potential energy is very low and the two charges are strongly attracted to one another. To see this, think about what must happen to pull two charges apart which have a very negative potential energy. If we want r to become very large, then in Coulomb's Law, we want V to get close to zero. If V is a large negative number, then we have to add a lot of energy to bring V up to zero. Therefore, a large negative value of V means that the two particles are strongly attracted to each other since it requires a lot of work to pull them apart. In the equation above, V will be a large negative number when several things are true: the charges must have opposite signs, so that multiplying them together gives a negative number. All this means is that opposite charges attract. The larger the charges, the stronger the attraction. In addition, r must not be large and preferably will be fairly small. These simple conclusions must be kept in mind. Two particles with large opposite charges close to one another must be strongly attracted to one another. The smaller the charges or the larger the distance, the weaker the attraction. In many ways, it is fair to say that Coulomb's Law forms the foundation of everything we know about the chemistry of atoms and molecules. Therefore, it is very important to understand the conclusions of the previous paragraph. Without them, we can make no further progress in our understanding of atoms.

4.3 Observation 1: Periodic Properties of the Elements We now have much more information about the dierences between the atoms of dierent elements. We know how many electrons and protons each atom contains, and we know where these charged particles are in the atom, with the protons in a very small nucleus and the electrons occupying the vast empty space around the nucleus. It seems that we should be able to account for the chemical properties of these atoms by using this information. However, we rapidly run into a surprising result. Remember that the atomic number tells us how many protons and electrons an atom contains. We observe that atoms with very similar atomic numbers often have very dierent chemical properties. For example, carbon's atomic number is 6 and nitrogen's is 7, so they have very similar numbers of protons and electrons. But as we have seen, elemental carbon is a solid and elemental nitrogen is a gas. Oxygen's atomic number is 8, just one greater than nitrogen, but oxygen reacts with most other elements, sometimes violently, whereas nitrogen is so unreactive that it is often used to provide an inert atmosphere to store chemicals. Also surprisingly, elements with very dierent atomic numbers can have quite similar chemical properties. The elements uorine and chlorine are both gases and both exist as diatomic molecules in nature, F2 and Cl2 . Both are highly reactive and will combine with hydrogen to form acids, HCl and HF. They both combine with metals like sodium and magnesium to form solid salts with similar molecular formulas, like NaF and NaCl. But their atomic numbers are quite dierent: F's atomic number is 9, and Cl's is 17. It should seem strange that elements with very dierent numbers of charged particles should behave alike chemically, but elements with very similar numbers of charged particles should behave very dierently chemically. There must be more to learn about what determines the chemistry of individual atoms. One clue is found in looking at the pattern of elements which have similar chemical properties. We can go to Table 4.1: Atomic Number and Relative Atomic Mass of Elements, which lists the elements by relative atomic mass and, as we now know, by atomic number. Let's pick out three elements with similar properties: lithium, sodium, and potassium. All three of these elements are soft metals with low melting points, all three react violently with water, and all three form salts with chlorine with the similar molecular formula, LiCl, NaCl, KCl. Because these elements are so similar, they can be regarded as a group, and a name has been given to this group, the alkali metals. Surprisingly, immediately before each of these elements in the list are three elements which are also very similar to each other but very dierent from the alkali metals. These Available for free at Connexions

33 are the noble gases, helium, neon, and argon. These elements are all gases to very, very low temperatures, and they are all very unreactive, sometimes called inert. The fact that each alkali metal is always preceded by a noble gas suggests that there is a pattern to the properties of the elements. We can see this again by looking at the elements immediately after each alkali metal. These are beryllium, magnesium, and calcium, and again these three have very similar properties. They are all soft metals with higher melting points than the alkali metals, and they all form salts with chlorine with similar molecular formulas, BeCl2 , MgCl2 , and CaCl2 . Because they are similar, we place them in a group together, which has been called the alkaline earth metals. In fact, we can keep this up with other groups. The elements that are just before the noble gases are uorine and chlorine, which again have very similar chemical properties as we discussed before. Later on in the list of elements, the noble gas krypton is immediately preceded by the element bromine, which also has similar properties to chlorine and uorine. We place uorine, chlorine, and bromine in a group together called the halogens.

Atomic Number and Relative Atomic Mass of Elements

Atomic Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Element Name Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminum Silicon Phosphorous Sulfur Chlorine

Element Symbol H He Li Be B C N O F Ne Na Mg Al Si P S Cl

Relative Atomic Mass 1.0079 4.0026 6.941 9.0122 10.811 12.011 14.007 15.999 18.998 20.180 22.990 24.305 26.982 28.086 30.974 32.066 35.453

continued on next page

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CHAPTER 4. ELECTRON SHELL MODEL OF AN ATOM 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton

Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

39.948 39.098 40.078 44.956 47.876 50.942 51.996 54.938 55.845 58.933 58.693 63.546 65.39 69.723 72.61 74.922 78.96 79.904 83.80

Table 4.1

The striking observation is that the groups of elements appear periodically in the ranking of the elements by atomic number. Looking at the list in Table 4.1: Atomic Number and Relative Atomic Mass of Elements, in each case we nd, in order, a halogen, a noble gas, an alkali metal, and an alkaline earth metal, and this pattern repeats itself over and over again. This observation is called the Periodic Law, and it is the reason that the usual table of the elements is called the Periodic Table, which is arranged with the elements in each group placed together in columns. Periodic Law: The chemical and physical properties of the elements are periodic functions of the atomic number. This observation is very surprising! To see this, consider an analogy. Imagine we looked for a pattern in the grades of students in a class by where they like to sit. We would not be surprised to nd a pattern. Perhaps, for example, the most attentive students sit in a group near the front of the class and make the highest grades. However, we would be very surprised if we were to rank the students in order of decreasing grades and discover that every tenth student (1st , 10th , 21st , 31st ) in the list sat in the rst row, every other tenth student (2nd , 12th , 22nd , 32nd ) sat in the second row, and so forth. That would be very unexpected and very hard to explain. But we would look for a reason for the pattern. In a similar way, we have seen a surprising pattern in the behavior of the elements with atomic number, and we must look for a reason for that pattern.

4.4 Observation 2: Ionization Energies of the Atom We are now ready to use Coulomb's Law, as discussed in the Foundation, to understand the attraction of the electrons ying about the nucleus to the positive charge of the nucleus. Recall that the attraction of two Available for free at Connexions

35 charges together depends on the sizes of the charges and the distance between them. The size of the charge on an electron is often called e. In the case of an atom with atomic number Z, there are Z protons so the nuclear charge is +Ze. The attraction of an electron at distance r away from this nucleus is given by the potential energy in Coulomb's Law: V (r) = (+Zer)(−e)

This means that an electron close to the nucleus would be more strongly attracted to the nucleus, because its potential energy is much more negative. A large negative potential energy means that we would have to add a lot of energy to the electron to remove it from the atom so that r could become large. This also means that an electron in an atom with a large atomic number, Z, would be more strongly attracted to its nucleus than an electron in another atom with a smaller atomic number. We can actually observe the attractions of the electrons to the nucleus by measuring the amount of energy required to remove the electron from the atom. This energy is called the ionization energy of the atom because it is the energy required to take a neutral atom and turn it into a charged ion: A (g) → A+ (g) + e- (g) In this chemical process, A is an atom, the (g) means that this atom is in the gas phase, and A+ (g) is the same atom with one electron removed, leaving behind a positive charge. We call A+ an ion. Think about how the ionization energy is related to the potential energy in Coulomb's Law. For an electron to be removed from the atom, r must become very large so that the potential energy becomes essentially zero. If an electron began with a negative potential energy V(r), we would have added at least this must energy to bring the potential energy up to zero. Therefore, a large negative V(r) would require a large ionization energy. We often call the ionization energy, IE, and it is typically measured in kiloJoules (kJ) per mole of atoms. At this point, we don't need to worry about how this experiment is actually done. The experimental data found from measuring the ionization energies of the elements are shown in the graph in Figure 4.1.

Figure 4.1: Ionization energy of the main group elements by atomic number.

The rst thing to notice in Figure 4.1 is that the Periodic Law denitely applies to this particular property of the atoms. As we look at the atoms with increasing atomic number, the ionization energies increase and then abruptly decrease, over and over. At the top of each peak sits a noble gas element. At the bottom of Available for free at Connexions

36

CHAPTER 4. ELECTRON SHELL MODEL OF AN ATOM

each valley sits an alkali metal. Between each alkali metal and the next noble gas, the ionization energies increase almost continuously, with only small breaks along the way. This is a great conrmation of the Periodic Law. In this case, we can actually start to understand the patterns in Figure 4.1 from thinking about Coulomb's Law. The easiest property to understand is that in each period in the gure, the ionization energy increases as we move from an alkali metal (like Li) to the next noble gas (like Ne). This makes sense in terms of Coulomb's Law, because the atomic number, Z, is increasing, so the nuclear charge is increasing, so the electrons are more strongly attracted to the nucleus as Z increases. If we follow this line of thinking, we might expect that the ionization energies would just constantly increase with increasing atomic number Z. But this is not observed. Instead, after each noble gas, the ionization energy drops very dramatically to the following alkali metal (as in going from Ne to Na). Looking at the equation for Coulomb's Law, how can it be that a larger value of Z from Na compared to Ne can result in a much, much smaller ionization energy? The answer is in the denominator of Coulomb's Law, the distance r. The only way for the ionization energy to get smaller when Z gets larger is for r to get larger as well. In fact, r has to get much larger since the ionization energy falls by such a large amount. This means that the electron which is ionized in Na is much farther from the Na nucleus than the electron in Ne which is ionized is from the Ne nucleus. Perhaps this means that r just increases smoothly as the atomic number gets larger. But the data do not show this at all. After Na, the ionization energies of the atoms continue to grow again with the increasing atomic number, Z, and the larger charge of the nucleus. Putting these two conclusions together, we can create a model of the atom that explains the Periodic Law and the Ionization Energies in Figure 4.1. In each period of the Periodic Table from an alkali metal to a noble gas, the outermost electrons are all about the same distance from the nucleus. Each additional electron in each atom in one period is added to this same layer of electrons, which we will call an electron shell. At the end of the period, the shell is apparently full and unable to accommodate any more electrons. The next added electron in the next period must be in a new shell, much farther from the nucleus. It is worth reviewing how this shell model of the atom's structure explains the pattern in the ionization energies in Figure 4.1. This is left as a discussion question. Beyond that, though, there is an explanation for the Periodic Law. Why, for example, do the alkali metals Li, Na, and K, have such similar properties? The answer from our shell model is that each of these atoms has a single electron in its outermost shell, even though they have very dierent numbers of electrons in total. That outermost electron is the one which must determine the chemical behavior of the atom. Similarly, F and Cl have very similar chemical properties because, from the data in Figure 4.1, we can tell that each of these types of atoms have 7 electrons in the outermost shell. Again, the number of electrons in the outermost shell of the atom appears to determine the atom's chemical and physical properties. Because this outermost shell is the most important shell, it is given the name valence shell, and the electrons in that shell are called valence electrons. The word valence means importance. The valence electrons are the most important electrons in each atom.

4.5 Observation 3: Successive Ionization Energies of the Atom There is another more direct way for us to observe the number of valence electrons in each atom and to show that they are all in about the same shell. We can attempt to remove each electron in the valence shell, one after the other, increasing the charge on the ion to higher and higher values. We have discussed the rst ionization energy, which is the energy to remove one electron. The second ionization energy is the energy needed to remove an electron from the positive ion to form an ion with a +2 charge. There are also third and fourth ionization energies and beyond: First ionization energy IE1 A (g) → A+ (g) + e- (g) Second ionization energy IE2 A+ (g) → A2+ (g) + e- (g) Third ionization energy IE3 A2+ (g) → A3+ (g) + e- (g) Fourth ionization energy IE4 A3+ (g) → A4+ (g) + e- (g) Available for free at Connexions

37 The experimental data observed when doing these successive ionizations is shown in Table 4.2: Successive Ionization Energies of the Atoms (kJ/mol):

Successive Ionization Energies of the Atoms (kJ/mol)

IE1 IE2 IE3 IE4 IE5 IE6 IE7

Na 496 4562 6912 9543 13353 16610 20114

Mg 738 1451 7733 10540 13630 17995 21703

Al 578 1817 2745 11575 14830 18376 23293

Si 787 1577 3231 4356 16091 19784 23783

P 1012 1903 2912 4956 6273 22233 25397

S 1000 2251 3361 4564 7013 8495 27106

Cl 1251 2297 3822 5158 6542 9458 11020

Ar 1520 2665 3931 5770 7238 8781 11995

Table 4.2

Let's analyze this data, looking for evidence of the valence shell and the number of valence electrons. First, for every atom listed, IE2 is always greater than IE1 , IE3 is greater than IE2 , and so forth. This makes sense when we remember that the negatively charged electrons in an atom repel one another. Once an electron has been removed from an atom, the remaining electrons will have lower energy and be harder to remove. Looking more closely, though, the increases in the ionization energies are not very constant. In Na, IE2 is greater than IE1 by a factor of almost nine, but IE3 is greater than IE2 by a fraction, and the same is true with the higher ionization energies. This means that the rst electron in Na is fairly easy to remove, but the second one is much harder. This means that the rst electron removed from Na is in the valence shell, far from the nucleus, but the second electron removed is closer in and more strongly attracted. The third electron removed is harder still to remove, but not much. This means that the third electron removed is in the same shell as the second one. Therefore, Na has only one valence electron. Now look at Mg. The rst electron doesn't require much energy to remove, although it is more than in Na. The second electron is harder again, but the real change is when we try to remove the third electron. Suddenly we see an increase of a factor of 5 in the ionization energy. This means that Mg has two relatively easily removed electrons, and therefore Mg has two valence electrons. Looking at the data in Table 4.2: Successive Ionization Energies of the Atoms (kJ/mol), we can simply count the number of valence electrons in each atom. (It is hard to do this for Cl and Ar, because it is hard to remove this many electrons from a single atom.) The shell model of the atom tells us how the electrons are arranged in each atom, but it does not tell us why. We don't know why the electrons can't all be added to a single shell, because we don't know why a shell seems to ll up. It is clear from the data that the number of valence electrons in each noble gas is the number needed to ll the valence shell. But we don't have a reason why this is true. These questions will require further observations and reasoning.

4.6 Review and Discussion Questions 1. Provide the experimental evidence that reveals the electrons in an atom are grouped into a valence shell and inner shell electrons. 2. State and explain the evidence that reveals the outer shell of each inert gas is full. 3. Why does the ionization energy for each successive ionization increase for every atom? Why is the increase from IE4 to IE5 in Si much larger than any of the other increases for Si? By John S. Hutchinson, Rice University, 2011 Available for free at Connexions

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CHAPTER 4. ELECTRON SHELL MODEL OF AN ATOM

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Chapter 5

Quantum Electron Energy Levels in an Atom 1

5.1 Introduction The electron shell model for the atom provides signicant answers to many of the most important questions about the properties of atoms. For many of the problems that chemists need to solve, we don't need more details about the structure of the atom than what we can gain from knowing the numbers of electrons in the valence shell, the size of the valence shell, and the charge on the nucleus. As perhaps the best example of this, the Periodic Law of the elements is easily understood from the repeating pattern of lling a valence shell successively and starting over with a new shell. Even though more advanced and detailed theories of atomic structure have come along since the electron shell model was introduced, chemists return to this simple model to understand the properties of elements and the structures and reactions of molecules. We shall come back and explore these applications of the electron shell model in later Concept Development Studies. For now, there are still some nagging questions about this simple model. What does it mean for two or more electrons in an atom to be in the same shell? We don't have a model for what a shell is, other than a set of electrons which appear to be at about the same distance from the nucleus. But this does not give a clear picture of what the electrons are doing. We have said that the electrons move in the empty space surrounding the nucleus, but we have not yet asked how they move or where they move. Without knowing that, we cannot really know why electrons have similar or dierent energies. Probably the most important unanswered question is why the shells ll up. The arrangement of elements into groups and the periodicity of chemical properties both depend on the idea that a shell is lled by a certain number of electrons. Why is there a limit on the number of electrons which can t into a shell? Looking at the number of elements in each period, the number of electrons which lls a shell depends on which shell is being lled. There are 8 elements from lithium to neon and from sodium to argon, telling us that 8 electrons will ll the valence shells in each of those sets of elements. However, there are 18 elements from potassium to krypton and from rubidium to xenon, telling us that 18 electrons will ll the valence shells in each of those sets of elements. In the cases of hydrogen and helium, only 2 electrons will ll their shell. What determines how many electrons can t in a shell? What is special about the numbers 2, 8, and 18? Why is there a limit at all? These may seem like questions about only technical details. But the power of the electron shell model rests on these details, so we should nd out the answers to these questions. 1 This content is available online at .

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CHAPTER 5. QUANTUM ELECTRON ENERGY LEVELS IN AN ATOM

40

5.2 Foundation In this study, we will assume that we know the postulates of the Atomic Molecular Theory and our measurements of relative atomic masses. We know that an element is composed of individual atoms with identical masses, and we know that the atoms of dierent elements have dierent masses, which have been measured. We will also assume that we know that structure of an atom, with a tiny, massive, positively charged nucleus surrounded by a much larger empty space in which electrons move. The positive charge on the nucleus is equal to the number of protons in the nucleus and, in a neutral atom, is also equal to the number of electrons moving about the nucleus. The number, called the atomic number, is unique for each type of atom. No two elements have the same atomic number, and amongst the naturally occurring elements, no atomic number is skipped: for every integer up to 118 we know an element with that atomic number. We will assume knowledge of the concept of electron shells as a means of understanding the Periodic Law, which describes the chemical and physical properties of the elements. From the previous Concept Development Studies, we shall also assume an understanding of Coulomb's Law to describe the interactions of the protons and electrons in an atom. And we shall also assume an understanding that light, or electromagnetic radiation, can be described in terms of the frequency of the light and that the frequencies emitted by a light source can be separated by using a prism or diraction grating. We will use each of these understandings as we probe into greater detail about the structures of atoms.

5.3 Observation 1: The Spectrum of Hydrogen In the Concept Development Study on Atomic Structure, we discussed the experimental method called spectroscopy. In general in a spectroscopy experiment, we look at the frequencies of light that an atom or molecule emits when energy is added to it in some way (commonly by heating it or placing it in an electric arc but possibly also by shining light on it). Alternatively, we can also look at the frequencies of light which the atom or molecule absorbs when we shine light with a range of frequencies on the atom or molecule. These two techniques are called emission spectroscopy and absorption spectroscopy, and which method is used depends on what properties we are trying to measure. In the simple case of an atom, say hydrogen, the frequencies of light which the atom will absorb are the same as the frequencies of light which the atom will emit. We can look at either one, and the set of frequencies we see experimentally is called the spectrum of the atom. Since hydrogen is the simplest atom, with only a single proton and a single electron, we'll look at hydrogen's spectrum rst. The spectrum of hydrogen contains frequencies of light corresponding to visible light, ultraviolet light, and x-rays. The visible spectrum of hydrogen, which is just the four frequencies hydrogen emits that can be seen by the human eye, is shown in Figure 5.1. There are many more frequencies in hydrogen's spectrum than the ones that we can see. Table 5.1: Hydrogen Spectrum shows a set of these frequencies.

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41

Figure 5.1: Diagram showing the visible emission spectrum of hydrogen corresponding to light at 410,

434, 486, and 656 nm wavelengths.

Hydrogen Spectrum Wavelength (nm) 94 95 97 103 122 410 434 486 656 955 1005 1094 1282 1875

Frequency (THz) 3197 3157 3083 2922 2466 731 691 617 457 314 298 274 234 160

Region(color) Ultraviolet Ultraviolet Ultraviolet Ultraviolet Ultraviolet Visible (violet) Visible (violet-blue) Visible (blue-green) Visible (red) Infrared Infrared Infrared Infrared Infrared

Table 5.1

There doesn't appear to be a pattern of any sort in the data in Table 5.1: Hydrogen Spectrum, just a collection of numbers. The rst important conclusion to remember, though, is that only these frequencies are observed. Hydrogen atoms will not emit radiation with a frequency between these numbers, for example. Given the innite possibilities for what the frequency of light can be, this is a very small set of numbers observed in the spectroscopy experiment. Available for free at Connexions

CHAPTER 5. QUANTUM ELECTRON ENERGY LEVELS IN AN ATOM

42

As it turns out, there is a pattern in the data, but it is quite hard to see. Each frequency in the spectrum can be predicted by a very simple formula, in which each line corresponds to a specic choice of two simple positive integers, n and m. If we pick any two small integers (1, 2, 3, . . .) for n and m, we can calculate a frequency ν , from the
equation: ν =R×

1 1 n2 − m2

This is called the Rydberg equation, and the constant R is called the Rydberg constant, which has value 3.29 ≥ 1015 sec-1 . This is a truly remarkable equation, both because it is so simple and because it is complete. Every choice of n and m will produce a value of ν which is observed experimentally. And every ν that is observed in the experiment corresponds to a choice of n and m. There are no exceptions, no simple choices of n and m which don't produce a frequency observed in the spectrum, and no frequency in the spectrum for which we can't nd simple choices of n and m. A very interesting question, then, is what n and m mean. They must have some physical signicance to the hydrogen atom, since they predict a physical property of the hydrogen atom. The spectrum alone does not provide any further insights, so we'll need additional observations. What about the spectrum of atoms of other elements? Experimentally, we nd that the atomic spectrum of every element consists of a limited set of frequencies which are observed. Each element has a unique atomic spectrum, which a set of frequencies which uniquely identify that element. However, the Rydberg equation only predicts the frequencies for the hydrogen atom spectrum. There does not exist a similarly simple formula for any other atom.

5.4 Observation 2: The Photoelectric Eect To understand the spectrum of hydrogen and other elements, we need to have a better understanding of the energy associated with electromagnetic radiation. To begin, we should be clear that electromagnetic radiation is a form of energy. There are some simple ways to see this in everyday life, including the fact that water left in sunlight will become hotter just as if it had been heated over a ame. It is also possible to generate electrical power from light, and it is even possible to push an object using energy absorbed from light. The question we now ask is, How much energy is contained in light? Even more specically, How is the energy contained in the light related to the frequency and intensity of the light source? We need a way to measure the energy. One way is via the photoelectric eect. When light is directed at a metal surface under the correct conditions, experimental observations show that electrons are ejected from the metal. These electrons can be collected to produce a usable electric current. There are a number of simple applications of the photoelectric eect, including in remote control devices or electric eye door openers. We already know that removing an electron from an atom requires energy, so it must be true that removing an electron from a metal requires energy, as well. In the photoelectric eect, the energy required is provided by the light source. We can vary the frequency of the light source and the intensity of the light source to determine under what conditions enough energy is provided to eject an electron from the metal. This could be detected by measuring the electric current produced. The more electrons ejected, the greater the current. We can also measure the energy of the ejected electrons. This is hard to do, but it is possible to measure the kinetic energy of these electrons as they are collected. Before looking at the experimental data, we can stop to make a prediction about the results. We know that more intense light must contain greater energy, since light from the sun can certainly warm something up faster than a ashlight. So we would expect that more intense light would produce more and faster electrons. What would be harder to predict is the eect of changing the frequency of the light. So rst, let's use a light source with a single frequency. This can be done by passing the light source rst through a prism or diraction grating. Then we can vary the intensity of the light. The rst surprising result is that, if we choose a frequency too low for our source, there are no electrons produced by the photoelectric eect. There is no current at all. We have to use a light source with at least a minimum frequency, ν 0 , called the threshold frequency, in order to observe electrons ejected. This is puzzling. Even when we apply a very intense light source, which presumably provides a lot of energy, no Available for free at Connexions

43 electrons are ejected unless we are using light of at least the minimum frequency. And conversely, even if the intensity is very low so that not much energy is provided, we do get electrons ejected if the light source has frequency above the threshold. This is not what we would expect. The value of the threshold frequency depends on which type of metal we shine the light source on. Let's stick with a single type of metal, and let's use a light source with frequency greater than the threshold, so ν> ν 0 . With this light source in place, we'll vary the intensity of the light and measure the current and the kinetic energy of the ejected electrons. The results are shown in Figure 5.2.

Figure 5.2:

The results from shining light on metal and measuring the photoelectrons ejected.

the photoelectric current,

ν

Φ

is

is the frequency of light, I is the intensity of incident light, and KE is the

photoelectron kinetic energy. (a) If the frequency of light is high enough, the number of photoelectrons emitted increases directly with the light intensity. (b) However, the energy of the photoelectrons does not depend on the light intensity. (c) For photoelectrons to be emitted, the light frequency must be greater than a threshold value. (d) If the light frequency is high enough, the energy of the electrons increases directly with the frequency.

In part (a), we see that, just as expected, a more intense light above the threshold frequency produces more electrons and a greater current. But in part (b), we see that the more intense light above the threshold frequency does not produce electrons with greater kinetic energy. This is not expected, since more energy would have been expected to produce more energetic electrons. To dig further into these surprising results, let's x the intensity of the light and vary the frequency of the source while observing the electric current produced and the kinetic energy of the electrons ejected. The results are also shown in Figure 5.2. First, in part (c), we see the threshold frequency clearly. If the frequency is below the threshold frequency, ν 0 , then no electrons are ejected and there is no current. This suggests that the frequency of the light is related to the energy of the light so that we need a high enough frequency to provide a high enough energy. But part (c) also shows that if the light source is above the threshold, then we get a xed current, no matter how high the frequency. This also seems strange if the higher frequency light is higher in energy. What happened to the added energy at high frequency? The answer is in part (d), which shows that the kinetic energy of the ejected electrons increases when we increase the frequency of the light source. In fact, it is a very simple relationship. Assuming the frequency is above the threshold, increasing the frequency of the light increases the kinetic energy of the electrons in direct proportion. To summarize the results in Figure 5.2, light which is high intensity but low frequency will not eject Available for free at Connexions

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CHAPTER 5. QUANTUM ELECTRON ENERGY LEVELS IN AN ATOM

electrons. Light which has suciently high frequency will eject electrons whether the intensity is low or high. Increasing the intensity gives us more electrons but not more energetic electrons. Increasing the frequency gives us more energetic electrons but not more of them. This is a very challenging puzzle, and an analogy helps to reveal the subtle answer. Imagine trying to knock pieces out of a wall by throwing objects at it. We discover that, no matter how many ping pong balls we throw, we cannot knock out a piece of the wall. On the other hand, only a single bowling ball is required to accomplish the task. This is because, in either case, the energy from either the ping pong balls or the bowling ball is provided to the wall in individual collisions. Each ball hits the wall individually, whether there are many of them or only one. The key to understanding our imaginary experiment is in discovering that, although there are many more ping-pong balls than bowling balls, it is only the impact of each individual particle with the wall which determines what happens and whether a piece of the wall is knocked loose. The results of this experiment are similar to the observations of the photoelectric eect: very little high frequency light can accomplish what an enormous amount of low frequency light cannot. If we use this analogy, we can reasonably conclude that the energy of the light is supplied in bundles or packets of constant energy, similar to the energy of the balls being provided in packets. In light, we will call the packets of constant energy photons. In the photoelectric eect, each photon or packet of light energy hits the metal surface individually and acts individually. How much energy does a photon contain? This is revealed by looking at Figure 5.2(d), which shows that the kinetic energy of the ejected electrons increases in direct proportion to the frequency provided that the frequency is above the threshold. We can conclude that the light supplies energy to the electron, which is proportional to the light frequency, so the energy of each photon is proportional to the frequency of the light. This now accounts for the observation that the frequency of the light source must be above the threshold frequency. For a photon to dislodge a photoelectron, it must have sucient energy, by itself, to supply to the electron to overcome its attraction to the metal. It does not get any help from other photons, just like a single ping-pong ball acts alone against the wall. Since each photon must have sucient energy and since the energy is proportional to the frequency, then each photon must be of a sucient minimum frequency. Increasing the intensity of the light certainly must increase the total energy of the light, since we observe this in everyday life. This means that the intensity of the light is proportional to the number of photons in the light but not the energy of each individual photon. Therefore, if the frequency of the light is too low, the photon energy is too low to eject an electron. Think again of the analogy: we can say that a single bowling bowl can accomplish what many ping-pong balls cannot, and a single high frequency photon can accomplish what many low frequency photons cannot. The important conclusion for our purposes is that light energy is quantized into packets of energy. The amount of energy in each photon is proportional to the frequency of the light. Einstein rst provided these conclusions, along with the equation which gives the energy of a photon of frequency ν E = hν where h is a constant called Planck's constant.

5.5 Observation 3: Quantum Energy Levels in Hydrogen Atoms Observation 1 showed us that only certain frequencies of light are emitted by hydrogen atoms. Observation 2 showed us that the energy of light is quantized into photons, or packets of energy, whose energy is proportional to the frequency of the light ν . We can now think about combining these two observations into a single observation about the hydrogen atom. When a hydrogen atom emits light, it must be emitting a photon of energy and is therefore losing energy. A hydrogen atom consists only of a nucleus and a single electron moving about that nucleus. The simplest (and perhaps only) way for the hydrogen atom to lose energy is for the electron to lose some of its energy. Therefore, when a hydrogen atom emits radiation of a certain frequency, it is emitting a photon of a specic energy, and therefore, the electron loses that same very specic energy. In the spectrum of hydrogen, only certain frequencies are emitted. That means that only certain amounts of energy loss are possible for the electron in a hydrogen atom. How can this be? Why can't an electron Available for free at Connexions

45 in a hydrogen atom lose any amount of energy? The answer becomes clearer by thinking of an analogy, in this case of walking down a staircase or walking down a ramp. When you walk down a ramp, you can change your elevation by any amount you choose. When you walk down a staircase, you can only change your elevation by xed amounts determined by the xed heights of the steps and the dierence in heights of those steps. The energy of an electron is like the height of each step on a staircase, not like the height on a ramp, since the energy can only be changed by certain specic amounts. This means that the energy of an electron in a hydrogen atom can only be certain specic values, called energy levels. In other words, the energy of a hydrogen atom is quantized. The Rydberg equation tells us what these energy levels are. Recall that every frequency emitted by a hydrogen atom is predicted by the simple equation:
ν=R×

1 1 n2 − m2

Each emitted frequency must correspond to a certain energy hν , and this energy must be the energy lost by the electron. This energy must therefore be the dierence between two electron energy levels in the hydrogen atom. Let's label the energy the electron starts with as Em, where m is just an index that tells us where the electron starts. Similarly, let's label the energy the electron nishes with as En , where n is just a dierent index. The electron loses energy equal to Em  En , and this must equal the photon energy emitted: hν = Em - En We should be able to compare these two equations, since both contain a dierence between two quantities that depend on two indices, m and n. Each energy of the electron might be given by an index n as En =−h×R× n12

If so, then the energy lost
by an electron in the second equation above would be Em -En =−h×R× m1 − n1 =hν This equation is the same as the Rydberg equation found experimentally. Therefore, we can conclude that, in a hydrogen atom, the energy of an electron can only be certain values given by an integer index n and equal to 2

2

En =−h×R× n12

This means that the electron in a hydrogen atom can only exist in certain states with certain energies. These states must therefore determine the motion of the electron in the atom. Interestingly, this state of the electron is characterized by an integer, n, which we will now call a quantum number since it completely determines the quantized energy of the electron. This discussion has only been about the hydrogen atom. These results also apply generally to all atoms, since all atoms display only specic frequencies which they emit or absorb. Since only certain frequencies can be emitted by each atom, only certain energy losses are possible, and only certain energy levels are possible in each atom. However, the equation above applies only to the energy of a hydrogen atom, since the Rydberg equation only describes the experimental spectrum of a hydrogen atom.

5.6 Review and Discussion Questions 1. The photoelectric eect demonstrates that radiation energy is quantized into packets or photons. Explain how and why this observation is of signicance in understanding the structure of atoms. 2. Explain how we can know that higher frequency light contains higher energy photons. By John S. Hutchinson, Rice University, 2011

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CHAPTER 5. QUANTUM ELECTRON ENERGY LEVELS IN AN ATOM

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Chapter 6

Electron Orbitals and Electron Congurations in Atoms 1

6.1 Introduction In looking at the structure of atoms to understand their properties, we have come up with two very powerful models. The rst of these, the electron shell model developed in Concept Development Study 4, is based on observing the Periodic Law and the ionization energies of atoms. This led us to a model of the atom in which the electrons in an atom can be grouped together into shells with the electrons in a single shell seem to have similar energies. The second model is the quantum energy level description we observed in Concept Development Study 5. We observed that electrons in hydrogen atoms can only have certain specic energies, called energy levels. These two models seem quite similar. In both cases, it appears that the arrangement of the electrons in an atom can be understood by looking at the energies that the electrons can have. But there are still many questions about how these two models are related. We developed the shell model by comparing the properties of dierent atoms, each with dierent numbers of electrons. We developed the quantum energy level model by looking mostly at hydrogen atoms, which each have only a single electron. We would like to know how the energy levels in a hydrogen atom are related to the electron shells in atoms with many electrons. Perhaps the electrons in a single shell are in the same energy level, but we have not shown that. Even if we assume it is true, we still do not know what determines how many electrons can be in a single level. Without that, we do not know why each shell has a limit on how many electrons can t. And if we don't know that, then we don't know why the properties of the elements are periodic. To answer these important questions, we need to understand what it means for an electron to move about a nucleus. In the process of studying this, we will nd that electrons move very dierently than the ways in which we are used to particles moving. The results are surprising but they provide a rm foundation for understanding the Periodic Law of the elements.

6.2 Foundation In this study, we will assume that we know the postulates of the Atomic Molecular Theory and our measurements of relative atomic masses. We will also assume that we know that structure of an atom, with a tiny, massive, positively charged nucleus surrounded by a much larger empty space in which electrons move. The positive charge on the nucleus is equal to the number of protons in the nucleus and, in a neutral atom, is also equal to the number of electrons moving about the nucleus. We will assume knowledge of the concept 1 This content is available online at .

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CHAPTER 6. ELECTRON ORBITALS AND ELECTRON CONFIGURATIONS IN ATOMS

of electron shells as a means of understanding the Periodic Law, which describes the chemical and physical properties of the elements. From the previous Concept Development Studies, we shall also assume an understanding of Coulomb's Law to describe the interactions of the protons and electrons in an atom. We will assume an understanding that light is an electromagnetic wave, meaning that, as it travels, the electric part of the light and the magnetic part of the light oscillate like a wave. And we shall also assume an understanding of the conclusions of our observations of the photoelectric eect. This means we understand that the energy of light is quantized into photons, or packets of energy, whose energy is proportional to the frequency of the light.

6.3 Observation 1: Electron waves and the uncertainty principle Our rst step in discovering how electrons move requires us to examine the results of an experiment seemingly far removed from the questions we are asking. These results concern something called diraction, which happens when two waves collide. When two particles collide, they may bounce o of each other, stick together, or cause each other to change paths. A wave collision is perhaps hard to think about, since waves don't exist in a single location. For this reason, we sometimes think instead of the overlap of two waves which come across each other. When two waves collide or overlap, their motions interfere with one another, meaning that the waves can add together or subtract from one another. This interference can be either constructive or destructive, depending upon how the waves add together. If the high points of both waves coincide in the same place, then the waves add together to give a bigger wave with a greater amplitude. If the high point of one wave adds to the low point of the other wave, then the waves cancel each other out. We can have a wave with smaller amplitude, and in some locations the interference results in zero amplitude for the wave. This is called a node. It is easy to see these kinds of wave interference in water waves when the waves hit a barrier and bounce back. The wave coming to the barrier and the wave leaving the barrier interfere with each other, and a beautiful pattern of high points and low points emerges. Figure 6.1 is an example of a diraction pattern.

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Figure

6.1:

Photograph of wave interference coming towards the South Island of New Zealand

2

(http://www.ickr.com/x/t/0093009/photos/brewbooks/309494512/ ).

A common way to observe interference with waves is to allow the wave to encounter an obstacle. For a light wave, this could be a small slit. Dierent pieces of the light wave encounter the slit at dierent points, deecting in varying directions rather than going straight through the slit. When light is passed through a series or grid of small slits, the deected light pieces can then interfere with one another either constructively or destructively, depending upon the angle at which the light approaches the grid. Since we can get both increased and decreased amplitude, we can see a beautiful diraction pattern, just like water waves and as seen in Figure 6.2. Since the grid can produce a diraction pattern, it is called a diraction grating. 2 http://www.ickr.com/x/t/0093009/photos/brewbooks/309494512/

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CHAPTER 6. ELECTRON ORBITALS AND ELECTRON CONFIGURATIONS IN ATOMS

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Figure

6.2:

Diraction

of

a

red

laser

(633

nm)

through

3

a

diraction

grating

of

150

slits

(http://en.wikipedia.org/wiki/File:Diraction_150_slits.jpg ).

The comparison between the diraction patterns of water waves and of light is very strong evidence to prove to us that light moves like a wave. Recall, however, that our earlier conclusion is that light behaves as a collection of energy packets, or photons. This means that light has some characteristics which are like particles and some which are like waves. Up to this point, we have assumed that electrons are simply particles, behaving essentially as billiard balls or planets. To test this assumption, we try reecting electrons o of a surface of a metal and looking at the pattern produced when the electrons return from their interaction with the surface. Since the metal consists of atoms, the metal surface looks to the incoming electron like a diraction grating, with grooves spaced one atom apart. As we can see in Figure 6.3, we observe in this experiment that the reected electrons produce a pattern very similar to that observed by diracted light. Certain angles of incidence and reection produce no reected electrons. These angles are alternated with angles with strong probability for reection of electrons. This is very strong evidence that electrons move as waves! 3 http://en.wikipedia.org/wiki/File:Diraction_150_slits.jpg

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Figure 6.3:

Electron diraction pattern from parallel beams of electrons taken using a Transmission

4

Electron Microscope (http://en.wikipedia.org/wiki/File : DifraccionElectronesMET.jpg).

The concept of wave-like motion for electrons has been very dicult to imagine or visualize. What does it mean for a particle to move like a wave? This is very subtle, and we will discuss it later in this study. But we can visualize it. Very recently, scanning tunneling microscopy (STM) has been used to take images that clearly reveal this wave-like character. The STM mechanism can be used to literally pick up and place metal atoms in specic arrangements on metal surfaces. For example, iron atoms have been arranged to form a closed circle on a copper surface. An image of the resultant structure then taken using the STM shows not only the ring of iron atoms but also conspicuous waves inside the ring, which result from the motion of electrons moving within the ring and reecting o of the walls formed by the iron atoms. Two of the original images taken at IBM of these so-called quantum corrals are shown in Figure 6.4. 4 http://en.wikipedia.org/wiki/File

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Figure 6.4:

The color-enhanced STM image on the left shows iron atoms in a stadium corral shape

on a copper surface.

The series of four STM images on the right show the making of a circular

corral of iron atoms on a copper surface.

The images were originally created by IBM Corporation

(http://www.almaden.ibm.com/vis/stm/corral.html)

Interpretation of the wave motion of electrons is a very complicated proposition, and we will only deal at present with a single important consequence, namely the uncertainty principle. A property of wave motion is that, unlike a particle, the wave does not have a denite position at a single point in space. By contrast, our everyday experience with particles is that the location of a particle is precise. We can look at something and determine where it is with a great deal of certainty. But our experiments tell us that the electron travels as a wave, and we cannot determine precisely the location of a wave. We must conclude that we cannot determine the precise location of an electron in an atom. This is, for our purposes, the uncertainty principle arising from the branch of Physics called quantum mechanics. Even though we cannot determine the precise location of an electron within an atom, we can make measurements of the location of the electron. With these measurements, we nd that each results in a dierent value for the location. Even though we can't pin the electron down, we can determine a probability distribution for where the electron is observed. This probability distribution is the most that we can know about the location and motion of an electron. It is extremely dicult to observe, but it can be determined by calculations from the eld of quantum mechanics. The postulates (or rules) of quantum mechanics cannot be deduced from our experimental observations, and the calculations are far beyond what we need to worry about here. For what we need in this study, our observations, such as electron diraction and the quantized energy levels for the electron in a hydrogen atom, are all consistent with the predictions of quantum mechanics. We will treat the predictions of quantum mechanics as the equivalent of experimental observations, conclusions that we can work with and build on.

6.4 Observation 2: Electron orbitals in hydrogen atoms Quantum mechanics tells us that the motion of the electron in a hydrogen atom is described by a function, often called the wave function or the electron orbital and typically designated by the symbol Ψ. The Available for free at Connexions

53 electron orbital is the best information we can get about the motion of the electron about the nucleus. For a particular position (x,y,z) in the space about the nucleus, quantum mechanics tells us that |Ψ|2 is the probability of observing the electron at the location (x,y,z). The uncertainty principle we worked out above tells us that the probability distribution is the most we can know about the electron's motion. In a hydrogen atom, it is most common to describe the position of the electron not with (x,y,z) but rather with coordinates that tell us how far the electron is from the nucleus, r, and what the two angles which locate the electron, θ and φ. We won't worry much about these angles, but it will be valuable to look at the probability for the distance of the electron from the nucleus, r. There isn't just one electron orbital for the electron in a hydrogen atom. Instead, quantum mechanics tells us that there are a number of dierent ways for the electron to move, each one described by its own electron orbital, Ψ. Each electron orbital has an associated constant value of the energy of the electron, En . This agrees perfectly with our earlier conclusions in the previous Concept Development Study. In fact, quantum mechanics exactly predicts the energy levels and the hydrogen atom spectrum we observe. The energy of an electron in an orbital is determined primarily by two characteristics of the orbital. The rst characteristic determines the average strength of the attraction of the electron to the nucleus, which is given by the potential energy in Coulomb's law. An orbital which has a high probability for the electron to have a low potential energy will have a low total energy. This makes sense. For example, as we shall see shortly, the lowest energy orbital for the electron in a hydrogen atom has most of its probability near the nucleus. By Coulomb's law, the potential energy for the attraction of the electron to the nucleus is lower when the electron is nearer the nucleus. In atoms with more than one electron, these electrons will also repel each other according to Coulomb's law. This electron-electron repulsion also adds to the potential energy, since Coulomb's law tells us that the potential energy is higher when like charges repel each other. The second orbital characteristic determines the contribution of kinetic energy to the total energy. This contribution is more subtle than the potential energy and Coulomb's law. As a consequence of the uncertainty principle, quantum mechanics predicts that, the more conned an electron is to a smaller region of space, the higher its average kinetic energy must be. Remember that we cannot measure the position of electron precisely, and we dene the uncertainty in the measurement as ∆x. This means that the position of the electron within a range of positions, and the width of that range is ∆x. Quantum mechanics also tells us that we cannot measure the momentum of an electron precisely either, so there is an uncertainty ∆p in the momentum. In mathematical detail, the uncertainty principle states that these uncertainties are related by an inequality: (∆x)(∆p)≥ 4πh This inequality reveals that, when an electron moves in a small area with a correspondingly small uncertainty ∆x, the uncertainty in the momentum ∆p must be large. For ∆p to be large, the momentum must also be large, the electron must be moving with high speed, and so the kinetic energy must be high. (We won't need to use this inequality for calculations, but it is good to know that h is Planck's constant, 6.62×10-34 J·sec. We have previously seen Planck's constant in Einstein's equation for the energy of a photon.) From the uncertainty principle we learn that the more compact an orbital is, the higher the kinetic energy will be for an electron in that orbital. If the electron's movement is conned to a small region in space, its kinetic energy must be high. This extra kinetic energy is sometimes called the connement energy, and it is comparable in size to the average potential energy of electron-nuclear attraction. Therefore, in general, an electron orbital provides an energy compromise, somewhat localizing the electron in regions of low potential energy but somewhat delocalizing it to lower its connement energy. What do these orbitals look like? In other words, other than the energy, what can we know about the motion of an electron from these orbitals? Quantum mechanics tell us that each electron orbital is given an identication, essentially a name, that consists of three integers, n, l, and m, often called quantum numbers. The rst quantum number n tells us something about the size of the orbital. The larger the value of n, the more spread out the orbital is around the nucleus, and therefore the more space the electron has to move in. n must be a positive integer (1, 2, 3, . . .), so the smallest possible n is 1. In a hydrogen atom, this quantum number n is the same one that tells us the energy of the electron in the orbital, En. The second quantum number, l, tells us something about the shape of the orbital. There are only a Available for free at Connexions

CHAPTER 6. ELECTRON ORBITALS AND ELECTRON CONFIGURATIONS IN ATOMS handful of orbital shapes that we nd in atoms, and we'll only need to know two of these for now. l is a positive integer or 0, and it must be smaller than the value of n for the orbital. For example, if n is 2, l must be less than 2, so l can be either 0 or 1. In general, l must be an integer from the set (0, 1, 2, ... n-1). Each value of l gives us a dierent orbital shape. If l = 0, the shape of the orbital is a sphere. Since the 54

orbital tells us the probability for where the electron might be observed, a spherical orbital means that the electron is equally likely to be observed at any angle about the nucleus. There isn't a preferred direction. Since there are only a few shapes of orbitals, each shape is given a one letter name to help us remember. In the case of the l = 0 orbital with a spherical shape, this one letter name is s. (As an historical note, s doesn't actually stand for sphere; it stands for strong. But that doesn't mean we can't use s as a way to remember that the s orbital is spherical.) Figure 6.5 is an illustration of the spherical shape of the s orbital.

Figure 6.5: Approximate probability distribution of the 1s orbital shown with axes to emphasize the

3-dimensional spherical shape.

For now, the only other shape we will worry about corresponds to l = 1. In this case, the orbital is not spherical at all. Instead, it consists of two clouds or lobes on opposite sides of the nucleus, as in Figure 6.6. This type of orbital is given the one letter name p, which actually stands for principal. Looking at the p orbitals in Figure 6.6, it is reasonable to ask what direction the two lobes are pointing in. If it seems that the lobes could point in any of three directions, that is correct. There are three p orbitals for each value of n, each one pointed along a dierent axis, x, y, or z. The third quantum number m gives each orbital a name allowing us to distinguish between the three orbitals pointing perpendicularly to each other. In general, the Available for free at Connexions

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m quantum number must be an integer between l and +l. When l = 1, m can be -1, 0 or 1. This is why there are three p orbitals.

Figure 6.6: Approximate probability distributions of p orbitals shown along the x, y, and z axes.

Although we will not worry about the shape of the l = 2 orbitals for now, there are two things to know about them. First, these orbitals are given the single letter name d. Second there are ve d orbitals for each n value, since l = 2 and therefore m can be -2, -1, 0, 1, or 2. Chemists describe each unique orbital with a name which tells us the n and l quantum numbers. For example, if n = 2 and l = 0, we call this a 2s orbital. If n = 2 and l = 1, we call this a 2p orbital. Remember though that there are three 2p orbitals since there are three m values (-1, 0, 1) possible. The motion of an electron in a hydrogen atom is then easily described by telling the quantum numbers or name associated with the orbital it is in. In our studies, an electron can only be in one orbital at a time, but there are many orbitals it might be in. If we refer to a 2p electron, we mean that the electron is in an orbital described by the quantum numbers n = 2 and l = 1. The n = 2 value tells us how large the orbital is, and the l = 1 value tells us the shape of the orbital. Knowing the orbital the electron is in is for now everything that we can know about the motion of the electron around the nucleus.

6.5 Observation 3: Photoelectron spectra and electron congurations So far, we have been concerned almost entirely with describing the motion of the electron in a hydrogen atom. Fortunately, this will be helpful in understanding the motion of electrons in all other atoms. The most important dierences between the hydrogen atom and all other atoms are the charge on the nucleus, the number of electrons, and the eects of the repulsions of the electrons from each other. Electron-electron repulsion is not important in a hydrogen atom since it contains only a single electron, but it is very important in all other atoms. To begin to understand the energies and orbitals for electrons in other atoms, we need more experimental information. We look at a new experiment called photoelectron spectroscopy. This form of spectroscopy is similar to the photoelectric eect we discussed in the previous concept study. We shine light on an atom and measure the minimum frequency of light which will ionize an electron from an atom. Remember that the frequency of light corresponds to a specic energy of the photons in that light. When the frequency of light is too low, the photons in that light do not have enough energy to ionize electrons from an atom. As we increase the frequency of the light, we nd the minimum frequency, or threshold, at which electrons begin to ionize. As we continue to increase the frequency, we nd that additional electrons are ionized at higher thresholds. These electrons are more tightly bound to the atom, requiring more energy and thus a Available for free at Connexions

CHAPTER 6. ELECTRON ORBITALS AND ELECTRON CONFIGURATIONS IN ATOMS

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greater frequency of light to ionize. By nding these higher thresholds for ionization, we can measure the ionization energy of not only the outermost electron but also of each electron in each orbital. With a higher frequency of light, there is sucient energy to ionize a number of dierent electrons, each with its own energy. The dierent types of electrons are distinguishable from each other by their kinetic energies when they are ionized. The more energy which is required to ionize the electron, the less energy is left over for the kinetic energy of the ionized electron. This means that we can look at the energies of all of the electrons in an atom, not just the electrons with the highest energy. The threshold ionization energies for the rst twenty elements are given in Table 6.1: Threshold Ionization Energies.

Threshold Ionization Energies

Element H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca

Ionization Energy Thresholds (MJ/mol)

1.31 2.37 6.26 11.5 19.3 28.6 39.6 52.6 67.2 84.0 104 126 151 178 208 239 273 309 347 390

0.52 0.90 1.36 1.72 2.45 3.12 3.88 4.68 6.84 9.07 12.1 15.1 18.7 22.7 26.8 31.5 37.1 42.7

0.80 1.09 1.40 1.31 1.68 2.08 3.67 5.31 7.79 10.3 13.5 16.5 20.2 24.1 29.1 34.0

0.50 0.74 1.09 1.46 1.95 2.05 2.44 2.82 3.93 4.65

0.58 0.79 1.01 1.00 1.25 1.52 2.38 0.42 2.9 0.59

Table 6.1

There is a lot of data in this table, so we should take it line by line and look for patterns. We note that there is a single threshold for hydrogen and helium. In hydrogen, this makes sense, because there is only a single electron. The lowest energy orbital is the 1s orbital, since n = 1 is the lowest value and only l = 0 is possible when n = 1. But helium has two electrons. Why is there only a single type of electron in helium? It must be the case that both electrons in helium are in the same orbital with the same energy. Also, the ionization energy for helium is about double the ionization energy of hydrogen. This makes sense when we remember that the charge on the nucleus is in Coulomb's law: doubling the charge should double the strength of the attraction of the electron to the nucleus. This very strongly indicates that the two electrons Available for free at Connexions

57 in helium are in the same orbital as the one electron in hydrogen. If we want to describe the helium atom, we could state that it has two 1s electrons. We adopt a shorthand notation for this, 1s2 , meaning that there are two electrons in the 1s orbital. This is called the electron conguration of helium. This is an extremely important conclusion because it tells us that we can use the electron orbitals for the hydrogen atom to describe the motions of electrons in other atoms. We now look at lithium and beryllium and notice that there are two ionization energies for each, meaning that there are two types of electrons in each atom. In lithium, this denitely means that the electron conguration is not 1s3 . Apparently, there cannot be three electrons in a 1s orbital, so the third electron must go into a higher energy orbital. The next orbital higher in energy would be either a 2s or a 2p orbital. So lithium must have two electrons in the 1s orbital and one electron in either the 2s or 2p orbital, and the electron conguration is either 1s2 2s1 or 1s2 2p1 . Whichever it is, it appears that beryllium will have a similar electron conguration 1s2 2s2 or 1s2 2p2 , since there are only two ionization energies for beryllium. And we already know that two electrons can be in the same orbital. But which is the correct electron conguration? To nd out, we look at boron and notice that suddenly there are three ionization energies. One of them is much larger than the other two, and the other two are fairly similar. Certainly the large ionization energy is due to two electrons in the 1s orbital. It seems probable that one of the other two ionization energies is for an electron in the same orbital as lithium or beryllium. The third one represents an electron in a new orbital. Apparently, we cannot put a third electron into whichever orbital beryllium has two electrons in. This is just what we saw before: there appears to be a fundamental principle that only two electrons can move in the same electron orbital. This principle is part of a more general principle called the Pauli Exclusion Principle, named after its discoverer. We can use this principle and the ionization energies to determine the electron congurations of lithium, beryllium, and boron. If beryllium's conguration was 1s2 2p2 , we could put a third electron in a 2p orbital, because there are three 2p orbitals, as we recall from above. This would mean that boron's electron conguration would be 1s2 2p3 , and there would be only two ionization energies. But this is not right: the data show that there are three ionization energies for boron. If beryllium's electron conguration were 1s2 2s2 , then the added electron in boron would have to go into a new orbital, and boron's electron conguration would be 1s2 2s2 2p1 . The data in Table 6.1: Threshold Ionization Energies for boron match this conguration. Notice that it appears that the 2s electrons and the 2p electron have very similar ionization energies. This makes sense, since both have the same n value. We will later explore the reason why they don't have exactly the same energy. Does this concept account for the ionization energies of the next several elements (carbon to neon)? In each of these elements, there are only three ionization energies, and just as in boron, one is very large and the other two are smaller and comparable in size. But there are six elements from boron to neon. How can it be that we can have six electrons in the 2p orbital? The answer is that there are three 2p orbitals, so two electrons can move in each orbital for a total of six 2p electrons. For example, the electron conguration of neon would be 1s2 2s2 2p6 . The next obvious step in the data in Table 6.1: Threshold Ionization Energies comes with sodium, where a fourth ionization energy is observed. This means that there are electrons in four dierent types of orbitals. If our reasoning above is correct, this makes sense. There is no room for a seventh electron to move in the 2p orbitals, so one electron in sodium must be in a higher energy orbital. The next higher energy orbital would be either 3s, 3p, or 3d, since n = 3 is the next lowest energy level. The ionization energies bear this out, since one of them is quite large (the 1s electrons) two of them are moderately sized (the 2s and 2p electrons), and one is much smaller (the n = 3 electron). Just by looking at the pattern of the ionization energies for the elements from sodium to argon, it should be clear that we have the same pattern as for lithium to neon. This means that the same argument must apply, and the 3s orbital must have an electron in sodium and must have two electrons in magnesium. Sodium's electron conguration must be 1s2 2s2 2p6 3s1 and magnesium's must be 1s2 2s2 2p6 3s2 . Our ionization energy data have provided us with three conclusions. First we conclude that two, and only two, electrons can move in the same orbital. Second, the electron conguration for each atom can be found by assigning electrons two at a time to each orbital in increasing order of energy, with the s orbital Available for free at Connexions

CHAPTER 6. ELECTRON ORBITALS AND ELECTRON CONFIGURATIONS IN ATOMS lower in the energy than the p orbital for each n value. This is sometimes called the aufbau principle, after 58

the German word meaning, roughly, built up. Third is a very exciting conclusion based on the rst two. Notice that when n = 2, there are four dierent orbitals to which we can assign a total of eight electrons. After those eight electrons have been assigned, no more electrons can be assigned to orbitals with n = 2. But we've seen this before! The number eight corresponds to our shell model of the atom in which only eight electrons could be in the second or third shell of each atom. Now we have an explanation of why the electron shells ll up. Each shell corresponds to a particular n value, and there are a limited number of electrons which can t into the orbitals in each shell. Once those orbitals have all been assigned two electrons, the shell is full, and any additional electrons must be assigned to a new shell at a higher energy. Looking at the pattern of the data in Table 6.1: Threshold Ionization Energies reveals to us the ordering of the orbitals in terms of increasing energy, at least for this set of atoms. The data show that the orbitals, in order of increasing energy, are: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s etc.

6.6 Observation 4: The importance of electron-electron repulsion There is a single detail missing from our description, which is of some importance to our discussion. Why is the 2s orbital lower in energy than the 2p orbital? This is something of a surprise. If we go back to look at the energy levels for an electron in a hydrogen atom, we will recall that the energy depends only on the n quantum number. This is the same for an electron in the 2s and 2p orbitals in a hydrogen atom. But in all of the atoms from lithium to neon, it is clear that the 2s orbital is lower in energy than the 2p orbital. Hydrogen is unique because it has only a single electron. As a result, there is no electron-electron repulsion to aect the energy. Apparently, when there is electron-electron repulsion, an electron in the 2s orbital has a lower energy than an electron in the 2p orbital. Therefore, the movement of the electron in the 2s orbital must produce a smaller amount of electron-electron repulsion than that produced by the movement of the electron in the 2p orbital. Electrons do repel each other. If they move in such a way as to maintain distance from one another, the amount of this repulsion is smaller. The 2s orbital must do a better job of this than the 2p orbital. It is very dicult to calculate the energy from electron-electron repulsion because we do not know, on average, how far apart the electrons typically are so we can't use Coulomb's law in any simple way. We need a simple model instead. One way to do this is to think of the shell structure of the atom. An n = 2 electron is farther from the nucleus than an n = 1 electron. In lithium, for example, the lone 2s electron is on the outside of both the nucleus and the two 1s electrons. These two 1s electrons form a negatively charged core which surrounds the nucleus. In lithium, the nuclear charge is +3 and the charge on the core electrons is -2, so the 2s electron feels an attraction to a +3 charge and a repulsion from a -2 charge. A simple way to look at this is as though the 2s electron feels a net charge from the core of +1. This is sometimes called the eective nuclear charge because it takes into account both the attraction to the charge on the nucleus and the repulsion from the negative charge from the core electrons. In one way of viewing this, the nuclear charge is partially shielded by the negative charge of the core electrons, reducing the attraction of the 2s electron to the nucleus. This argument would seem to apply perfectly well to an electron in either the 2s or 2p orbital, since both are in the n = 2 shell. So this does not account for the dierence in the energies of these orbitals. It does account for the similarity of the energies of these orbitals, as clearly seen in the data in Table 6.1: Threshold Ionization Energies. But we still need to know why there is a dierence. The answer lies in remembering that in each orbital we only know the probabilities for where the electrons might be observed. Therefore, we cannot denitely state that the n = 2 electrons are outside of the 1s core electrons; we can only say that they probably are. As a result, the shielding eect is not perfect. In fact, there is some probability that an n = 2 electron might be found closer to the nucleus than the 1s electrons. This is called core penetration. When an n = 2 electron does penetrate the core, it is no longer shielded from the nucleus. In this case, the n = 2 electron is very strongly attracted to the nucleus and its energy is Available for free at Connexions

59 thus lowered. Therefore, the eect of core penetration is to reduce the shielding eect and therefore increase the eective nuclear charge. What is the extent of this penetration? The answer is in Figure 6.7, which shows the probability of nding an electron a distance r away from the nucleus for each of the 1s, 2s, and 2p orbitals. If we compare the probabilities, we see that there is a slightly greater probability (though still quite small) for the 2s electron to penetrate the core than for the 2p electron to do so.

Figure 6.7: Probability for an electron at a distance r from a hydrogen nucleus.

As a result of the core penetration, an electron in a 2s orbital feels a greater eective nuclear charge than just the core charge, which was approximated by assuming perfect shielding. Thus, the eective nuclear charge for a 2s electron is greater than the eective nuclear charge for a 2p electron. Therefore, the energy of an electron in the 2s orbital in beryllium is lower than it would be in the 2p orbital. Knowing how the electrons move about the nucleus explains the ionization energy data in Table 6.1: Threshold Ionization Energies and also explains the shell model of the atom by explaining why each shell can hold a limited number of electrons. This means that we can understand the Periodic Law and the trends in the Periodic Table using this new model. It turns out that we can also use our understanding of the motions of the electrons in atoms to help understand how and why atoms bond together to form molecules. This is a very important question for chemists, because understanding bonding is the key to understanding how and why molecules react with each other. Our understanding of electron orbitals will be very useful in answering these questions.

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CHAPTER 6. ELECTRON ORBITALS AND ELECTRON CONFIGURATIONS IN ATOMS 6.7 Review and Discussion Questions 60

1. Electron anity is the energy released when an electron is attached to an atom. If an atom has a positive electron anity, the added electron is attracted to the nucleus to form a stable negative ion. Why doesn't a Beryllium atom have a positive electron anity? Explain how this demonstrates that the energy of a 2s orbital is less than the energy of a 2p orbital. 2. Why does an inert gas atom have a high ionization energy but a low electron anity? Why do these properties combine to make the atoms of inert gases unreactive? 3. Consider electrons from two dierent subshells in the same atom. In photoelectron spectroscopy, the lower energy electron has a higher ionization energy but is observed to have lower kinetic energy after ionization. Reconcile the lower energy with the higher ionization energy with the lower kinetic energy. 4. Chlorine atoms have 5 distinct ionization energies. Explain why. Predict the number of ionization energies for Bromine atoms, and explain your answer. (Hint: examine the structure of the periodic table.) 5. Why does a Bromine atom have a much smaller radius than a Potassium atom, even though a Br atom has 16 more electrons than does a K atom? 6. Explain why electrons conned to smaller orbitals are expected to have higher kinetic energies. 7. Dene "shielding" in the context of electron-electron repulsion. What is the signicance of shielding in determining the energy of an electron? How is the aected by core penetration? By John S. Hutchinson, Rice University, 2011

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Chapter 7

Covalent Bonding and Electron Pair Sharing 1

7.1 Introduction The previous concept studies have provided us a detailed model for the structure and energetics of each atom of each element. This model is extremely useful in helping us understand how the chemical and physical properties of the elements are related to the properties of the individual atoms. Chemistry, though, is mostly about molecules and how they react with one another, so we now turn out attention to understanding molecules. We should begin by reviewing what we know so far. Recall that there are over 50 million known compounds in our world, each made up from less than roughly 90 commonly occurring elements. From the atomic molecular theory, we understand what it means to form a compound from its component elements. A compound consists of identical molecules, with each molecule made up of the atoms of the elements in a simple whole number ratio. We call this ratio of atoms the molecular formula, and from our work on mass ratios, we also know how to determine what the molecular formula is for any compound of interest. From our work on chemical algebra, we can quantify chemical reactions, determining how much product can be produced from a given amount of reactant. However, there are a great many fundamental questions about molecules we have not addressed or answered. Perhaps most interestingly, although we know what the molecular formula is for any compound, we don't know what determines the numbers of atoms which combine to form a molecule. Some combinations are observed (e.g. H2 O, H2 O2 ) and others are never observed (e.g. H6 O, HO6 ). We need to understand the principles which govern what combinations will work to form stable molecules and what combinations will not. In order to answer these questions, we will need to develop an understanding of the forces which hold molecules together. Since atoms are neutral, the forces cannot simply be attractions of oppositely charged atoms. We know that there are diatomic molecules like H2 and O2 . Why would identical atoms attract each other? Our knowledge of the charges contained inside atoms will be very helpful in understanding forces which bond atoms together. These forces must also be essential in determining the reactivity of a molecule, since these forces will determine how readily the atoms can be separated and recombined with atoms from other molecules. Predicting the reactivity of a substance is one of the great powers of Chemistry. If we wish to predict chemical reactivity of a particularly substance, we must understand the forces which bond atoms together in a molecule. 1 This content is available online at .

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7.2 Foundation In this study, we will assume that we know the postulates of the Atomic Molecular Theory and our measurements of relative atomic masses. We know that an element is composed of individual atoms with identical masses, and we know that the atoms of dierent elements have dierent masses, which have been measured. From these masses, we can determine the molecular formula of any substance or compound of interest. As such, we'll assume that these are known. We further assume the structure of the atom as a massive, positively charged nucleus, whose size is much smaller than that of the atom as a whole, surrounded by a vast open space in which negatively charged electrons move. These electrons can be eectively partitioned into a core and a valence shell, and it is only the electrons in the valence shell which are signicant to the chemical properties of the atom. The number of valence electrons in each atom is equal to the group number of that element in the Periodic Table. We will base much of our work on understanding the Periodic Law, which states that the chemical and physical properties of the elements are periodic functions of the atomic number. Finally, we will assume an understanding of Coulomb's Law, which describes the attractions and repulsions amongst charged particles.

7.3 Observation 1: Valence and the Octet Rule To begin to understand chemical bonding, we will examine the valence of an atom, which is dened as the atom's most common tendency to form bonds to other atoms. We can gure these out by looking at some common molecular formulae for molecules formed by each atom. We'll start with the easiest case, the atoms of the noble gases. Since these atoms do not tend to combine with any other atoms, we will assign their valence as 0, meaning that these atoms tend to form 0 bonds. This doesn't really get us very far. To nd the valence of an atom which does form bonds, let's pick molecules which contain only a single atom of the type we're interested in and see how many other atoms it can combine with. Oxygen is a good place to start. For example, a single O atom will combine with two H atoms to form the most common molecule H2 O. Only under rare circumstances would we nd any other combination of H and O in a neutral molecule. As such, it appears that the valence of an O atom is 2. Next we consider hydrogen, which combines with virtually any other element except the noble gases. Compounds containing hydrogen can contain a huge variety of the number of H atoms. However, molecules with a single H atom most typically contain only a single other atom, for example HF. A single C atom can combine with four H atoms, but a single H atom typically does not combine with more than one other atom. We do not typically see molecules like C4 H. A conspicuous feature of molecules containing hydrogen is that there are typically many more hydrogen atoms than other atoms. For example, hydrogen in combination with carbon alone can form CH4 , C2 H6 , C8 H18 , and many others. These observations lead us to conclude that an H atom has a valence of 1, meaning that a single H atom will typically only form 1 bond to another atom. This seems reasonable, since each H atom contains only a single proton and a single electron. This conclusion also is consistent with our conclusion that O atoms have a valence of 2, since the most common hydrogen-oxygen molecule is H2 O. We can use hydrogen's valence of 1 to nd the valence of other atoms. For example, the valence of C must be 4, since one C atom can combine with 4 H atoms, but not 5, and typically not 3. Nitrogen atoms have a valence of 3, to form NH3 . Fluorine atoms have a valence of 1, to form HF molecules. This concept also applies to elements just below carbon, nitrogen, oxygen, and uorine. Silicon will form SiH4 , so an Si atom has a valence of 4. Phosphorous forms PH3 , so P has a valence of 3, and Sulfur forms H2 S, so S has a valence of 2. Each halogen atom (Cl, Br, I) prefers to form molecules by combining with a single hydrogen atom (e.g. HCl, HBr, HI), so each halogen has a valence of 1. We can make further progress using the valence of the halogens. Lithium, sodium, potassium, and rubidium each bind with a single Cl atom to form LiCl, NaCl, KCl, and RbCl. Therefore, they also have a valence of 1. Because we also nd that, for example, the combination of two potassium atoms with a single oxygen atom forms a stable molecule, our assignments are all still consistent, since oxygen's valence of 2 can be satised by the two K atoms, each with a valence of 1. We can proceed in this manner to assign a valence to each element by simply determining the number of atoms to which this element's atoms prefer to bind. Available for free at Connexions

63 If we arrange the valences according to Periodic Table as in Table 7.1: Most Common Valence of Each Element in Periods 2 and 3, we discover that there is a pattern. Just as we would expect from the Periodic Law, elements in the same group all share a common valence.

Most Common Valence of Each Element in Periods 2 and 3 Li 1 Na 1

Be 2 Mg 2

B 3 Al 3

C 4 Si 4

N 3 P 3

O 2 S 2

F 1 Cl 1

Ne 0 Ar 0

Table 7.1

The inert gases with a valence of 0 sit to one side of the table. Each inert gas is immediately preceded in the table by one of the halogens: uorine precedes neon, chlorine precedes argon. And each halogen has a valence of one. This one step away, valence of one pattern can be extended. The elements just prior to the halogens (oxygen, sulfur, selenium, tellurium) are each two steps away from the inert gases in the table, and each of these elements has a valence of two (e.g. H2 O, H2 S). The elements just preceding these (nitrogen, phosphorus, antimony, arsenic) have valences of three (e.g. NH3 , PH3 ), and the elements before that (carbon and silicon most notably) have valences of four (CH4 , SiH4 ). The two groups of elements immediately after the inert gases, the alkali metals and the alkaline earths, have valences of one and two, respectively. Hence, for many elements in the periodic table, the valence of its atoms can be predicted from the number of steps the element is away from the nearest inert gas in the table. This systemization is quite remarkable and is very useful for remembering what molecules may be easily formed by a particular element. Next we discover that there is an additional very interesting aspect to the pattern of the valences: for elements in Groups 4 through 8 (e.g. carbon through neon), the valence of each atom plus the number of electrons in the valence shell in that atom always equals eight. For instance, carbon has a valence of 4 and has 4 valence electrons; nitrogen has a valence of 3 and has 5 valence electrons; oxygen has a valence of 2 and has 6 valence electrons. We have made one of the most important observations in Chemistry, the Octet Rule: Octet Rule: For elements in Groups 4 through 8 in Periods 2 and 3 of the Periodic Table, the valence of each atom plus the number of valence electrons in each atom is equal to 8.

7.4 Observation 2: Covalent Bonding As a way to think about this pattern, remember that for each of the atoms in these two periods, the valence shell can accommodate eight electrons. It turns out that the valence of each atom in Groups 4 through 8 is equal to 8 minus the number of valence electrons the atom has. For example, the valence of an N atom is 3, which is equal to 8 minus the number of valence electrons in N, which is 5. This tells us that, for each of these atoms, the number of bonds the atom typically forms is equal to the number of vacancies in its valence shell. This suggests a model which would account for the Octet Rule. It appears that each atom in these Groups attempts to bond to other atoms so as to completely ll its valence shell with electrons. For elements in Groups 4 through 8, this means that each atom attempts to complete an octet of valence shell electrons. (Why atoms should behave this way is a question unanswered by this model.) Consider, for example, the combination of two chlorine atoms to form Cl2 . Each chlorine atom has seven valence electrons and seeks to add a single electron to complete an octet. Hence, chlorine has a valence of 1. Either chlorine atom could satisfy its valence by taking an electron from the other atom, but this would leave the other atom now needing two electrons to complete its valence shell. The only way for both atoms to complete their valence shells simultaneously is to share two electrons. Each atom donates a single electron Available for free at Connexions

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to a shared electron pair. This sharing of electrons is what we call a chemical bond. More specically, we call this a covalent bond, so named because the bond acts to satisfy the valence of both atoms. The two atoms are thus held together by the need to share the electron pair. Let's apply this to a molecule with an H atom, HCl. H atoms are not expected to have an octet. In fact, the valence shell in H can accommodate just two electrons. Because an H atom has 1 valence electron, it thus has a single vacancy, or a valence of 1. Our model for electron pair sharing works for H atoms as well. When H and Cl bond, they share a pair of electrons forming a covalent bond so that the H atom has a full valence shell (two electrons) and the Cl atom has a full valence shell (eight electrons). We now have two important pieces of our model. First, atoms form a covalent bond by sharing a pair of electrons. Second, for atoms in Groups 4 through 8, they tend to share enough electrons to ll the valence shell with 8 electrons, and this determines the number of covalent bonds they form. An H atom shares one pair of electrons to form a single covalent bond.

7.5 Observation 3: Molecular Structures in Compounds of Carbon and Hydrogen Many of the most important chemical fuels are compounds composed entirely of carbon and hydrogen, i.e. hydrocarbons. The smallest of these is methane CH4 , a primary component of household natural gas. Other simple common fuels include ethane C2 H6 , propane C3 H8 , butane C4 H10 , pentane C5 H12 , hexane C6 H14 , heptane C7 H16 , and octane C8 H18 . There is a very interesting consistency in these molecular formulae: in each case, the number of hydrogen atoms is two more than twice the number of carbon atoms, so that each compound has a molecular formula like Cn H2n+2 . (Try it out!) This suggests that there are strong similarities in the valences of the atoms involved which should be understandable in terms of our valence shell electron pair sharing model. Since each H atom can only bond to a single other atom, the carbon atoms in each molecules must be directly bonded together. In the easiest example of ethane, the two carbon atoms are bonded together, and each carbon atom is in turn bonded to three hydrogen atoms. This would t our model of valence, since each carbon atom is bonded to four other atoms (three hydrogens and the other carbon). By sharing an electron pair with each of those four atoms, each carbon atom lls its valence shell with eight electrons. This example was not dicult. In most other cases, it is not so trivial to determine which atoms are bonded to which. This is because there may be multiple possibilities which satisfy all the atomic valences. As the number of atoms and electrons increases, it may also be dicult to determine whether each atom has an octet of electrons in its valence shell. We need a system of counting the valence electrons which makes it easy for us to see these features more clearly. To start, we create a notation for each atom which displays the number of valence electrons in the unbonded atom explicitly. In this notation, carbon and hydrogen look like

Figure 7.1

where the dots represent the single valence electron in hydrogen and the four valence electrons in carbon. Note that the C atom valence electrons are all unpaired. This is because we know that the valence of a C atom is four, so there are four valence electrons available to be shared with other atoms. Available for free at Connexions

65 Using this notation, it is now fairly easy to represent the shared electron pairs and the carbon atom valence shell octets in methane and ethane. For each pair of bonded atoms, we share an electron pair from the valence shell electrons. This gives for methane and ethane:

Figure 7.2

Recall that each shared pair of electrons represents a chemical bond. These drawing are examples of what are called Lewis structures, after G.N. Lewis who rst invented this notation. These structures reveal, at a glance, which atoms are bonded to which, so we call this the structural formula of the molecule. There are two things to check about the electrons in the structural formula. First, we cannot have created or lost any valence electrons. For example, in ethane we started with four valence electrons from each carbon and one valence electron from each hydrogen, for a total of 14 electrons. The structural formula of ethane drawn above has 14 valence electrons, so that is correct. Second, if we have satised the valence of each atom, each carbon should have an octet of electrons and each hydrogen should have two electrons. We can also easily count the number of valence shell electrons around each atom in the bonded molecule and verify that this is also correct. In a larger hydrocarbon, the structural formula of the molecule is generally not predictable from the number of carbon atoms and the number of hydrogen atoms, because there may be more than one possible arrangement. In these cases, the molecular structure must be given to deduce the Lewis structure and thus the arrangement of the electrons in the molecule. However, with this information, it is straightforward to create a Lewis structure for molecules with the general molecular formula Cn H2n+2 such as propane, butane, etc. For example, the Lewis structure for normal butane (with all carbons linked one after another) is found to be:

Figure 7.3

There are no hydrocarbons where the number of hydrogen atoms is greater than two more than twice the number of carbons. For example, CH5 does not exist, nor does C2 H8 . If we try to draw Lewis structures for CH5 or C2 H8 which are consistent with the octet rule, we will nd that there is no way to do so. Similarly, CH3 and C2 H5 are observed to be so extremely reactive that it is impossible to prepare stable quantities of either compound. And again we will nd that it is not possible to draw Lewis structures for these molecules Available for free at Connexions

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which obey the octet rule. We come to a very important and powerful conclusion: when it is possible to draw a Lewis structure in which each carbon has a complete octet of electrons in its valence shell, the corresponding molecule will be stable and the hydrocarbon compound will exist under ordinary conditions. After working a few examples, it is apparent that this always holds for compounds with molecular formula Cn H2n+2 .

7.6 Observation 4: Double and Triple Bonds in Compounds of Carbon and Hydrogen Although our model so far does a good job of describing molecules with the formula Cn H2n+2 , there are many stable hydrocarbon compounds with molecular formulae in which the number of hydrogen atoms is less than 2n+2. Simple examples are ethene C2 H4 and acetylene C2 H2 in which there are not enough hydrogen atoms to permit each carbon atom to be bonded to four atoms each. In each molecule, the two carbon atoms must be bonded to one another. When we arrange the electrons so that the carbon atoms share a single pair of electrons and then attach hydrogen atoms to each carbon, we wind up with rather unsatisfying Lewis structures for ethene and acetylene:

Figure 7.4

Note that, in these structures, neither carbon atom has a complete octet of valence shell electrons, but these are both stable compounds. We need to extend our model to work for these types of molecules. These structures indicate that the carbon-carbon bonds in ethane, ethene, and acetylene should be very similar, since in each case a single pair of electrons is shared by the two carbons. However, we can observe that the carbon-carbon bonds in these molecules are very dierent chemically and physically. First, we can compare the energy required to break each bond (the bond energy or bond strength). Second, it is possible to observe the distance between the two carbon atoms, which is referred to as the bond length. Bond lengths are typically measured in picometers (1 picometer (pm) = 10-12 m).

Properties of Carbon-Carbon bonds in Two Carbon Atom Molecules Ethane Ethene Acetylene (Ethyne) Bond strength (kJ/mol) 347 589 962 Bond length(pm) 154 134 120 Table 7.2

Looking at the data, it is very clear that the bonding between the carbon atoms in these three molecules must be very dierent. The bond strength increases when there are fewer H atoms, and the bond length gets progressively shorter with fewer H atoms. These observations are reinforced by looking at carbon-carbonbond Available for free at Connexions

67 strengths and lengths in other Cn H2n+2 molecules, like propane and butane. In all of these molecules, the bond strengths and lengths are comparable to those in ethane. The bond in ethene is about one and a half times stronger than the bond in ethane; this suggests that there might be a second bond between the two carbon atoms. This could be formed from the two unpaired and unshared electrons in the ethene structure above. Similarly, the bond in acetylene is about two and a half times stronger than the bond in ethane, so we can imagine that this results from the sharing of three pairs of electrons between the two carbon atoms. These assumptions produce the following Lewis structures:

Figure 7.5

These structures appear to make sense from two regards. First, these structures would explain the trend in carbon-carbon bond strengths as arising from the increasing number of shared pairs of electrons. Second, each carbon atom has a complete octet of electrons. Our model now reveals that there is a double bond in ethane and a triple bond in acetylene. We thus extend our model of valence shell electron pair sharing to conclude that carbon atoms can bond by sharing one, two, or three pairs of electrons as needed to complete an octet of electrons, and that the strength of the bond is greater when more pairs of electrons are shared. Moreover, the data above tell us that the carbon-carbon bond in acetylene is shorter than that in ethene, which is shorter than that in ethane. We conclude that triple bonds are shorter than double bonds, which are shorter than single bonds.

7.7 Observation 5: Hydrocarbon Compounds Containing Nitrogen, Oxygen, and the Halogens Many compounds composed primarily of carbon and hydrogen also contain some oxygen or nitrogen, or one or more of the halogens. These include the majority of biomolecules, such as amino acids and proteins. This means that it would be very desirable to extend our understanding of bonding by developing Lewis structures for these types of molecules. Recall that a nitrogen atom has a valence of 3 and has ve valence electrons. In our notation, we could draw a structure in which each of the ve electrons appears separately in a ring, similar to what we drew for C. However, our new understanding of covalent bonding tells us that unpaired electrons on atoms are shared with other atoms to form bonds. Having all ve valence electrons in an N atom unpaired would imply that an N atom would generally form ve bonds to pair its ve valence electrons. Since the valence is actually 3, our notation should have three unpaired electrons. One possibility looks like:

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Figure 7.6

Note that this structure leaves three of the valence electrons unpaired and thus ready to join in a shared electron pair. The remaining two valence electrons are paired, and this notation implies that they therefore are not generally available for sharing in a covalent bond. This notation is consistent with the available data, i.e. ve valence electrons and a valence of 3. Pairing the two non-bonding electrons seems reasonable in analogy to the fact that electrons are paired in forming covalent bonds. We can draw similar structures for oxygen and uorine. The other halogens will have structures like F, since they have the same valence and the same number of valence electrons.

Figure 7.7

With this notation in hand, we can now analyze structures for molecules including nitrogen, oxygen, and the halogens. The hydrides are the easiest:

Figure 7.8

Note that the octet rule is clearly obeyed for oxygen, nitrogen, and the halogens. At this point, it becomes very helpful to adopt one new convention: a pair of bonded electrons will now be more easily represented in our Lewis structures by a straight line, rather than two dots. Double bonds and triple bonds are respectively represented by double and triple straight lines between atoms. We will continue to show non-bonded electron pairs explicitly with two dots. For example, ethanol has the molecular formula C2 H6 O. The two carbon atoms are bonded together and the oxygen atom is attached to one of the two carbons; the hydrogen atoms are arranged to complete the valences of the carbon atoms and the oxygen atom: Available for free at Connexions

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Figure 7.9

In this structure, each line connecting two atoms represents a shared pair of electrons, or a covalent bond. The non-bonding pairs on oxygen are often called lone pairs. It is important for us to include them in our structure for three reasons. First, including the lone pairs helps us check that we have drawn a structure with the correct number of valence electrons. Let's check this for this drawing. Each carbon atom contributes four valence electrons, the oxygen atom contributes six, and each hydrogen atom contributes 1. There are thus a total of 2(4)+6+6(1)=20 valence electrons. Counting electrons in the drawing, there are eight covalent bonds, each of which represents two valence electrons, and two lone pairs, for a total of 20 valence electrons. Second, drawing the lone pairs helps us see that the octet rule is obeyed for the O atom. Third and perhaps most importantly, we will later learn that lone pairs of electrons are important in determining the physical and chemical properties of molecules.

7.8 Review and Discussion Questions 1. Compounds with formulae of the form Cn H2n+2 are often referred to as "saturated" hydrocarbons.Using Lewis structures, explain how and in what sense these molecules are saturated. 2. Molecules with formulae of the form Cn H2n+1 (e.g. CH3 , C2 H5 ) are called "radicals" and are extremely reactive. UsingLewis structures, explain the reactivity of these molecules. 3. State and explain the experimental evidence and reasoning which shows that multiple bonds are stronger and shorter than single bonds. 4. Compare N2 to H4 N2 . Predict which bond is stronger and explain why.

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Chapter 8

Molecular Structures

1

8.1 Introduction We have developed a model by which we can understand molecular formulas. In the previous concept development study, we began by asking what determines which combinations of atoms will form stable molecules and compounds and why some combinations are never observed. We combined our knowledge of the electronic structures of atoms with our knowledge of molecular formulas and the common valences of atoms to develop the octet rule. The rule tells us that the most common valence of each main group atom is equal the number of spaces for electrons in the valence shell of the atom, which is eight minus the number of valence electrons. Thus, according to the octet rule, atoms tend to bond such that they have eight valence electrons in the bonded molecule. The Lewis structure model implements the octet rule and reveals which atoms are bonded to which other atoms in a molecule and whether these bonds are single, double, or triple bonds. Thus, Lewis structures provide additional information about molecules that we cannot learn just from molecular formulas. This model of molecular structure, like most good scientic models, poses at least as many new questions as the questions it was designed to answer. What new information can be gleaned from our knowledge of the structure of a molecule? It seems reasonable for us to assume that the properties of a compound are related to the properties of the individual molecules of the compound. It also seems reasonable to assume that the properties of individual molecules are related to their structures. If we could identify the properties of molecular structures which are related to their chemical or physical properties, we could understand these properties and perhaps even design molecules which have interesting or valuable properties. This is one of the very most important questions in chemistry. Before we can pursue it, we need more detail and insight about molecular structures. In this concept development study, we will examine various arrangements of atoms in a variety of molecules and look for common structures. We will nd both variety and common features. We will also examine how the electrons are arranged in these molecular structures. This will allow us to interpret molecular structures accurately.

8.2 Foundation We will assume that we know the valences of the common main group elements, and from this, we know the octet rule. This assumes knowledge of the valence shell model of the electronic structure of atoms. We know that we can combine these atoms in the Lewis structure model to build molecular structures which satisfy the valences of the main group elements and t the octet rule for valence electrons. We have found that molecules which have Lewis structures that t the octet rule typically form stable compounds. And we 1 This content is available online at .

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found that, for combinations of atoms where we cannot construct Lewis structures that t the octet rule, the compounds are either unstable or non-existent. By comparing our molecular structures to experimental data, we observed that double bonds are stronger and shorter than single bonds, and triple bonds are stronger and shorter than double bonds.

8.3 Observation 1: Compounds with Identical Molecular Formulas In simple molecules, such as CH4 , it is fairly clear that the molecule consists of a single carbon atom surrounded by and bonded to four hydrogen atoms. There really is only one way that the molecule could be structured. But even in molecules only slightly larger than this, we discover some interesting observations. Let's consider the molecular formula C2 H6 O. Experimentally, we discover that there are two dierent compounds with the molecular formula C2 H6 O, and these two compounds have very dierent physical and chemical properties. One of the compounds is a liquid at room temperature with a boiling point of 78 C. The other is a gas with a boiling point of -25 C. This might seem surprising in a couple of ways. First, how could two compounds have the same molecular formula? We might have expected that a specic combination of the elements would produce a specic compound, but this is clearly not true. And second, even if we do imagine more than one possible compound, we would most probably have guessed that they would be similar in their properties since they contain the same elements in the same proportions. How can two compounds with the same molecular formula have so very dierent properties? To understand these observations, we can use our model for molecular structures. From our earlier work, it is clear that the two carbon atoms and the oxygen atom must be bonded together, surrounded by hydrogen atoms. But in what order are they bonded? One option would be C-C-O, and a second would be C-O-C. The rst of these would give us the molecular structure:

Figure 8.1

The second would give us:

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Figure 8.2

By all of our criteria for the Lewis model of molecular structures, both of these look right. Each carbon atom and the oxygen atom in each structure show the proper valence and a complete valence shell. There are no unpaired electrons in either structure. These two molecular structures must be consistent with our experimental observation that there are two compounds with this molecular formula. Each of the compounds must correspond to one of the dierent molecular structures. Which one is which? Determining which molecular structure corresponds to which compound is an interesting experimental problem. We'll explore this further in the next section. For now, we'll assume that it is experimentally possible to determine which is which. In this case, the rst molecular structure above, containing the C-O-H arrangement, is the liquid at room temperature and is called ethanol or ethyl alcohol. The second molecular structure, containing the C-O-C arrangement, is a gas at room temperature and is called dimethyl ether. Compounds with the same molecular formula are called isomers of one another. In the example above, dimethyl ether is an isomer of ethanol and vice versa. It is very common to observe isomers for any given molecular formula. Some of these are simple to see. Let's consider the molecular formula C2 F4 Cl2 . Each halogen atom generally has a valence of 1 with seven valence electrons. This suggests that the two carbon atoms should be bonded together, as in ethane, with the six halogen atoms arranged three per carbon. But there is more than one way to do this:

Figure 8.3

In one molecular structure, the two Cl atoms are on the same carbon atom. In the other, the two Cl atoms are separated on dierent carbon atoms. We might guess that this dierence in arrangements does Available for free at Connexions

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not matter, since in both cases, the F and Cl atoms are all bonded to the carbon atoms, and the carbon atoms are bonded together. Our experimental observations prove this guess is wrong. There are two isomers of C2 F4 Cl2 , one with a boiling point of 3 C and a melting point of -57 C and the other with a boiling point of 3.8 C and a melting point of -94 C. Dierences in the arrangement of atoms in similar molecules clearly do matter, even if those dierences don't seem all that great. In some cases, isomers have very obviously dierent structures. Let's look at the various isomers with the molecular formula C5 H10 . Perhaps the simplest structure is one with a double bond between two of the carbon atoms. However, in a chain of ve carbon atoms, there are two dierent places where the double bond might be. Each of these corresponds to a dierent compound:

Figure 8.4

The ve carbon atoms don't have to be lined up in a single chain:

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Figure 8.5

One isomer of C5 H10 which might not have been obvious is a structure in which the ve carbon atoms form a ring:

Figure 8.6

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CHAPTER 8. MOLECULAR STRUCTURES

All seven of these isomers are dierent compounds with distinct physical and chemical properties. From these and many similar observations we can conclude as a general rule that isomers have dierent molecular structures, which give rise to the diering properties of the compounds.

8.4 Observation 2: Molecular Properties and Functional Groups The previous conclusion leads us to a new question: what is it about the dierences in molecular structure that produces the dierences in properties? This is a huge question which, in many ways, is one of the fundamental questions of the eld of Organic Chemistry. Although we cannot hope to provide all the answers to this question in this one study, we can study a few examples to develop the fundamental concept. We begin by looking for common connections amongst molecular structures and molecular properties. From the observations we just discussed, it seems that every arrangement of the atoms in a molecule produces properties which are unlike the properties of any other molecular structure. Let's go back and examine the two isomers of C2 H6 O, ethanol and dimethyl ether. As noted above, ethanol is a liquid at room temperature, and is completely soluble in water. Dimethyl ether is a gas at room temperature and is less soluble in water than ethanol. The most notable dierence between the two molecular structures is that the oxygen atom is bonded to a hydrogen atom in ethanol. There are no O-H bonds in dimethyl ether. The structure C-O-H appears in a large number of molecules. And experimentally, we nd that molecules with the C-O-H group have similar molecular properties to ethanol. All are liquids at room temperature and most are soluble in water. These compounds are, as a class, called alcohols. These observations lead us to conclude that common properties are due to the common C-O-H group. The C-O-H is called the hydroxyl group. When we nd a group of atoms which gives a specic set of properties, or function, to molecules, we call that group of atoms a functional group. Using this new term, we would say that the class of molecules called alcohols contain the hydroxyl functional group. There are many functional groups found in chemistry, each of which gives specic properties to the class of molecules containing it.

Functional Group Alkane Alkene

Functional Structure

Group Functional Group Amine

Functional Structure

Aldehyde

continued on next page

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Group

77 Alkyne

Ketone

Arene

Amide

Alcohol

Carboxylic Acid

Ether

Ester

Table 8.1

As mentioned above, the eld of Organic Chemistry is in large part about nding ways to synthesize new molecules containing specic functional groups which therefore form compounds with specic properties we might want. Much of Biochemistry can be understood by looking at the functional groups present in biomolecules.

8.5 Observation 3: Molecular Structures with more than one Bonding Structure Earlier in this concept study, we learned that dierent molecules can be formed from the same set of atoms. Isomers are molecules with the same molecular formula but dierent molecular structures, and are therefore distinctly dierent compounds. Rearranging the atoms in the molecule creates a new compound with new physical and chemical properties. What if, instead of rearranging the atoms in the molecule, we rearrange the electrons? This would possibly change where double or triple bonds are located and whether there are lone pairs of electrons or not. Does this also give rise to new compounds and therefore new isomers? We rst look at benzene, a compound with the molecular formula C6 H6 . For six carbon atoms, there are not very many hydrogen atoms. Compare benzene to hexane, which has the molecular formula C6 H14 . This means that there must be several double or triple carbon-carbon bonds in benzene. Experiments reveal to us two facts about the molecular structure of benzene. First, the six carbon atoms are arranged in a ring, not a chain, and each carbon atom is bonded to a single hydrogen atom. Second, the bonds between the carbon atoms in the ring all have the same length as one another. The second observation tells us that, somehow, all of the bonds in benzene are identical to each other. The correct molecular structure which explained these observations was a puzzle for chemists. A structure in which the six carbon atoms are arranged in a ring would be: Available for free at Connexions

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CHAPTER 8. MOLECULAR STRUCTURES

Figure 8.7

This structure cannot be the right structure though. Although it satises the octet rule and the valences of the carbon atoms, it does not correctly explain why the bonds are all the same length. This structure would predict three shorter bonds and three longer bonds. A clue to the right structure is found from the value of the bond length, 139 pm. This length is between the typical length of a single bond, 153 pm, and the typical length of a double bond, 134 pm. This is a confusing clue: it suggests that the bonds in benzene are neither single bonds nor double bonds. We clearly need to expand on our model of Lewis structures. If we look at the molecular structure proposed above, we can see that we made an arbitrary choice of where to put the double bonds. To satisfy the valences of the carbon atoms, we need to alternate the double bonds and single bonds, but we could have chosen the other three C-C bonds to be double:

Figure 8.8

Available for free at Connexions

79 The dierence between these two structures is not the arrangement of the atoms, but rather the arrangement of the electrons. Are these isomers? If they are, then there are two benzene compounds. Experimentally, we nd only one. So these two structures must not be isomers. Rearranging the electrons in a molecular structure does not produce new compounds. What then are we to do with the two structures drawn above? There is no reason why one would be preferred over the other, so both must be equally correct for benzene. But neither of them alone is correct because each of them predicts bond lengths which do not match experimental observations. One possible answer is that there is a single molecular structure for benzene which combines both structures together. This means that the double bonds in benzene are not xed in one set of locations or the other. Rather, the bond lengths tell us that the double bonds are spread out around the six carbon ring uniformly. The language that chemists use to describe this phenomenon is that the correct structure of benzene is a hybrid of the two structures drawn above. The word hybrid refers to something that contains properties of more than one element. Benzene has a single molecular structure that combines the properties of both of the above structures at the same time. How does this explain the experimental bond lengths? If we look at each C-C bond and combine the properties of the two structures, each bond has the properties of a single bond and of a double bond. This means that the bond length would be somewhere between a single bond and a double bond, and this is just what is found experimentally. A Lewis structure which represents this hybrid is:

Figure 8.9

The drawing on the left uses dotted lines to represent the double bonds which are neither here nor there but are rather delocalized around the six carbon ring. The drawing on the right is another way to represent this idea, but the solid ring represented the double bonds is somewhat easier to draw and therefore more commonly used by chemists. Chemists refer to the delocalization of the electrons as resonance, and the structure above is often called a resonance hybrid. This concept applies to a number of molecules. A good example is ozone, O3 . Experiments show that the two O-O bond lengths are equal, 128 pm. We can compare this to the double bond length in O2 , which is 121 pm, and to the single bond length in hydrogen peroxide, H-O-O-H, which is 147 pm. From our model, we might conclude that the O-O bonds in O3 are partially double and partially single, just like in benzene. How would our model account for this? Available for free at Connexions

80

CHAPTER 8. MOLECULAR STRUCTURES We can draw two equivalent Lewis structures for ozone:

Figure 8.10

Based on our observations and our model, we can conclude that the correct molecular structure of ozone is a resonance hybrid of these two structures in which the double bond is delocalized over both O-O bonds.

8.6 Interpretation of Lewis Structures Before further developing our model of chemical bonding based on Lewis structures, we pause to consider the interpretation and limitations of these structures. At this point, we have observed no information regarding the geometries of molecules. For example, we have not considered the angles measured between bonds in molecules. Consequently, the Lewis structure model of chemical bonding does not at this level predict or interpret these bond angles. Therefore, although the Lewis structure of methane is drawn as

Figure 8.11

this does not imply that methane is a at molecule, or that the angles between C-H bonds in methane are 90 ◦ . Rather, the structure simply reveals that the carbon atom has a complete octet of valence electrons in a methane molecule, that all bonds are single bonds, and that there are no non-bonding electrons. Similarly, one can write the Lewis structure for a water molecule in two apparently dierent ways:

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81

Figure 8.12

However, it is very important to realize that these two structures are identical in the Lewis model because both show that the oxygen atom has a complete octet of valence electrons, forms two single bonds with hydrogen atoms, and has two pairs of unshared electrons in its valence shell. Drawing the structure either way does not convey any dierent information. Neither of the structures is more right or more wrong. In the same way, the following two structures for Freon 114

Figure 8.13

are also identical. These two drawings do not represent dierent structures or arrangements of the atoms in the bonds. Finally, we must keep in mind that we have drawn Lewis structures strictly as a convenient tool for our understanding of chemical bonding and molecular stability. It is based on commonly observed trends in valence, bonding, and bond strengths. However, these structures must not be mistaken as observations themselves. As we encounter additional experimental observations, we must be prepared to adapt our Lewis structure model to t these observations, but we must never adapt our observations to t the Lewis model.

8.7 Review and Discussion Questions 1. Draw molecular structures for all isomers for the molecular formula C6 H14 . Available for free at Connexions

82

CHAPTER 8. MOLECULAR STRUCTURES 2. Draw Lewis structures for molecules with the molecular formula N2 O. Are these structures dierent isomers? Or are they resonance structures of the same molecule? How do experimental observations dier for structures which are isomers versus structures which are resonance structures? 3. A student drew the following six structures for C4 H10 . Which of these are correct structures? Of the correct structures, which are identical and which are isomers? Are any of the structures resonance structures?

Figure 8.14

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83 4. We drew two benzene structures with alternating double bonds as follows:

Figure 8.15

It looks like the structure on the right is simply a 60 ◦ rotation of the structure on the left. Viewed this way, it might seem that these two are the same structure and the correct structure of benzene is just one of these. Provide experimental evidence and reasoning to demonstrate that this is not the correct interpretation of these two Lewis structures. 5. Amino acids are small molecules which link together in long chains to form proteins. The name amino acid comes from the fact that each molecule contains an amine group and a carboxylic acid group (see the table of functional groups in this study.) The amino acid alanine has molecular formula C3 H7 NO2 . Given that it is called an amino acid, use the table to draw a molecular structure for alanine.

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Chapter 9

Energy and Polarity of Covalent Chemical Bonds 1

9.1 Introduction The Lewis model of chemical bonding is one of the most powerful models in all of Chemistry. With the simple concepts of the octet rule and the sharing of an electron pair to form a chemical bond, we can predict what combinations of atoms and which molecular structures are likely to be stable or unstable. For example, we can predict that the molecular formula C4 H10 should correspond to a stable molecular structure, since it possible to draw a Lewis structure in which all four carbon atoms have an octet of valence electrons and each hydrogen atom has two valence electrons. In fact, we can even predict that there should be two such molecular structures corresponding to two dierent stable compounds, butane and methyl propane (often called isobutane). Lewis structures also allow us to predict the relative strengths and lengths of chemical bonds. For example, we can predict that in the molecule propene C3 H6 , one of the C-C bonds is a single bond and the other is a double bond; correspondingly, we can predict that the double bond is shorter and stronger than the single bond. In general, for any molecule containing some combination of atoms of C, N, H, O, and any of the halogens (F, Cl, Br, or I), we can predict with condence that the molecule will be a stable compound if we can draw a Lewis structure in which the C, N, O, and halogen atoms all have an octet of valence electrons. This is a very signicant statement, since the overwhelming majority of the molecules of organic chemistry are composed of only these atoms. The converse is also generally true: if we cannot draw a Lewis structure in which these atoms have an octet of valence electrons, we will predict fairly condently that a stable compound of that molecular formula probably does not exist. Note that stable does not mean non-reactive. In this context, stable means that the compound exists and can be isolated for long periods of time. Methane CH4 is a stable compound which can be manufactured, captured, stored, and transported. However, it is also a reactive compound, reacting rapidly with oxygen gas in one of the most common combustion reactions in the world. These are the reasons why we say that the Lewis structure model of covalent bonding is one of the most important models in all of Chemistry. However, like most good models, the understanding we get from the model comes with even more questions we might want to ask. In this Concept Development Study, we will address several of these, while others will wait for later studies. First, why does sharing a pair of electrons create a chemical bonding? More pointedly, what does it even mean to share an electron between two atoms? And why do atoms prefer to share a pair of electrons, rather than one or three? Of course, we've also seen that atoms can share more than one pair of electrons, creating double or triple bonds. We might have guessed that, with more electrons being shared, these electrons would repel each other, leading to a 1 This content is available online at .

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86

higher energy, but this is not the case. Why does sharing more than one pair of electrons produce a stronger bond? Once we have an understanding of what sharing electrons means, we will also ask whether the atoms really do share these electrons fully. We already know that the energies of electrons in dierent atoms are quite dierent. We have observed and analyzed the ionization energies of electrons in atoms. These results showed that the electrons in some atoms such as uorine, for example, are much more strongly attracted to their nuclei than in atoms such as carbon, for example. Knowing this, we might readily ask whether the electrons shared by a C atom bonded to an F atom are actually shared equally. If they are not equally shared, does it matter? Does this aect the physical or chemical properties of the molecules? As always, we will examine experimental observations to help us understand the answers to these questions.

9.2 Foundation In this Concept Development Study, we will assume that we already know the basic rules of the Lewis model of chemical bonding. Chemical bonds between atoms consist of one, two, or three pairs of shared electrons, respectively resulting in single, double, or triple bonds. Atoms in groups IV, V, VI, and VII, most importantly including C, N, O, and F, share electrons in pairs such that, in stable molecules, these atoms typically have eight electrons in their valence shells. An H atom will share one pair of electrons with other types of atoms to form stable molecules. We will need Coulomb's Law to understand how the bonding electrons interact with the nuclei of the atoms that share them. We have already used Coulomb's Law to understand the energies of electrons in atoms and how these vary from one type of atom to another. These same lines of reasoning will be useful for electrons in molecules. Though we will not need all of the postulates and conclusions of Quantum Mechanics, there are a few that we will call on to answer the questions posed above. This makes sense, since any discussion of where electrons might be found or how they might move certainly requires us to take a quantum mechanical view of electrons. Most notably, we'll need to recall that the motion of electron is described by an orbital which provides the probability for where the electron might be. In addition, we'll recall that an electron can exist in two dierent spin states, and that two electrons' motions may be described by the same orbital only if they have dierent spin states.

9.3 Observation 1: The Simplest Chemical Bond, H2 +

Let's start with the easiest molecule we can imagine. We need at least two atoms of course, so that means we need two nuclei. The smallest nucleus is just a single proton, in other words, a hydrogen ion, H+ . We could try to build a molecule formed from just two of these, but this seems unlikely. The two positively charged nuclei would just repel each other, so there can't be such a molecule. We need at least one electron to provide some attraction. Is one enough? The best way to answer that question is to ask whether the combination of two H+ nuclei plus one electron is a stable molecule, H2 + . Experimentally, it is indeed possible to observe H2 + . Even though it is far less common than H2 , it does exist. What does this mean in terms of chemical bonding? It means that H2 + does not spontaneously fall apart into an H atom and an H+ ion. Instead, to break the bond we would have to do some work, or, in other words, add some energy. This amount of energy can be measured and is 269 kJ/mol. That is actually quite a lot of energy which must be added to pull an H atom and an H+ ion apart, so a strong bond is formed when the two H+ nuclei share a single electron. This might seem surprising, since our model of chemical bonding based on Lewis structures regards a chemical bond as a sharing of a pair of electrons. We'll examine the importance of sharing two electrons instead of one in the next section. But rst, let's ask why sharing an electron creates a bond. A better way to ask this question would be why is the energy of the electron lower when the electron is shared by two nuclei instead of being near to only Available for free at Connexions

87 one? We know that an electron and a proton attract each other via Coulomb's law, and the closer the two are, the lower the energy. As such, if an electron can be attracted to two protons, it has a lower energy even if the two protons are not next to each other. This is illustrated in Figure 9.1 (Attractions and Repulsions in H2 + )b, which shows the attraction of the electron to both nuclei as well as the repulsion between the two nuclei. When the electron is near to two nuclei, it has a lower potential energy than when near to only one because its attraction to a second positively charged nucleus further lowers its energy. We can rearrange the electron and the two nuclei, such that the electron is on the outside of the molecule (Figure 9.1 (Attractions and Repulsions in H2 + )a). Here, it is close to one nucleus but not the other. In this case, the repulsion of the two nuclei is greater than the attraction of the electron to the distant nucleus. Just using Coulomb's law, the energy of the electron in this position is slightly lower than when it is on one nucleus, but the repulsion of the nuclei is large. From a comparison of the arrangements of the electron and the two nuclei, it appears that we get a lowering of energy when the electron is between the two nuclei and not otherwise. This suggests what sharing an electron means when forming a chemical bond. Since the shared electron has a lower energy, we would have to raise the electron's energy to pull the two nuclei away from each other. Now we know why a chemical bond is formed when two atoms share at least one electron. The energy of the electron is lower when shared in the molecule than it is in the separated atoms because it is attracted to both positive nuclei at the same time. To break the bond formed by sharing the electron, we must do work, which means we have to add energy to raise the energy of the electron to be able separate the atoms.

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Attractions and Repulsions in H2+

2+

Figure 9.1: Attractions and Repulsions in H

For now, we should make a note to ourselves that all of this discussion of electron energy includes only a discussion of potential energy. It also assumes that the electron is at some particular location where we can calculate its potential energy. This ignores the uncertainty principle of quantum mechanics, suggesting that the model is not physically accurate. Both of these issues need to be corrected, so we need a better understanding of the sharing of electrons in chemical bonds.

9.4 Observation 2: The Simplest Molecule, H2

The hydrogen molecule is familiar to us from our early eorts to determine molecular formulae. Avogadro was the rst person to suggest that hydrogen gas consists of diatomic molecules, H2 , instead of individual hydrogen atoms. This was a perplexing idea for the chemists of the early nineteenth century. Why would identical H atoms be attracted to each other? Of course, they did not know anything about the structure of these atoms, including that each atom contains a positive nucleus and an electron. From our work on H2 + , we now have a clue as to what holds the two H atoms together. It must be electron sharing. What observation can we make about the bonding in H2 ? Clearly, it is a stable molecule. Although H2 molecules are highly reactive, they do not spontaneously fall apart into H atoms except under very extreme Available for free at Connexions

89 circumstances. Experimental data tell us that the bond energy of H2 is 436 kJ/mol. This is even more energy than is required to break the bond in H2 + , nearly twice as much in fact. Remember that to break the bond in H2 + we must raise the energy of the shared electron. Perhaps, in H2 , we have to raise the energy of the two electrons, costing us about twice the energy. This seems to be a good starting point for understanding the strong bond in H2 . But there are some troubling questions about this simple picture. This model would suggest that sharing more than two electrons should give an even greater bond energy. But this is only true for some molecules, and it is certainly not true for H2 . There is something signicant about sharing a pair of electrons, rather than one or three. In addition, the strength of the bond (436 kJ/mol) is actually less than double the strength of the bond with one shared electron (269 kJ/mol). There must be another factor at work. And nally, there are the questions we ended with in the last section: what about kinetic energy? And what about the uncertainty principle, which states that an electron is not actually localized? Let's work our way backwards through these questions. We know from our study of quantum mechanics that the motion of an electron is described by an orbital which provides the probability for where the electron might be found. We can't actually know where an electron is, but we can look at its probability distribution. This is true for electrons in molecules just like it is in atoms. We just need to observe what an orbital looks like for an electron that is shared by two nuclei. Such a molecular orbital for the electron in H2 + is shown in Figure 9.2 (Bonding Molecular Orbital). We've seen images like this when we discussed atomic orbitals. Remember that this image of a cloud gives us a probability: where there are many dots, the probability is high. The electron can be found near either nucleus but can also be found with high probability in the area between the two nuclei. This seems encouraging when thinking about the energy of the electron in the molecule as we discussed in the previous section. When the electron is in the region between the two nuclei, the potential energy of the electron is lower.

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CHAPTER 9. ENERGY AND POLARITY OF COVALENT CHEMICAL BONDS

90

Bonding Molecular Orbital

Figure

9.2:

orbital

image

courtesy

2

of

M.

Winter,

The

Orbitron

website

(http://winter.group.shef.ac.uk/orbitron/ )

There is a second interesting aspect to the molecular orbital. It is larger in space than the atomic orbital that the electron would occupy if it were in a single H atom. Why does this matter? Thinking back to our study of the energies of electrons in atomic orbitals, we can recall that both potential energy and kinetic energy are important. The kinetic energy of the electron is easiest to understand by remembering the uncertainty principle. The more the electron is conned, the less certain is its momentum. This means that, on average, the electron moves faster with greater kinetic energy when it is conned to a smaller space. When less conned, the electron can have lower kinetic energy. Since the molecular orbital connes the electron less than the atomic orbital does, the kinetic energy of the electron is lower in the molecular orbital than it is in the atomic orbital. (Some very complicated calculations show that this fact is the single most important factor in lowering the energy of the electron in a bond. For simplicity, we will assume that both lower potential energy and lower kinetic energy contribute to the strength of a chemical bond.) Figure 9.2 (Bonding Molecular Orbital) shows a molecular orbital for H2 + . However, we were discussing the bond strength in H2 and, in particular, the fact that two electrons seem to be better than one. Again, we have to refer back to our study of the quantum energy levels and electron congurations of atoms. Remember that there were rules from both quantum and experimental observations which restrict what energy states electrons can occupy. Most importantly, there is an important aspect about two electrons: only two electrons can occupy a single orbital. As a very good model then, we can imagine that the two electrons in H2 both move as described by the probability in the molecular orbital in Figure 9.2 (Bonding Molecular Orbital). 2 http://winter.group.shef.ac.uk/orbitron/

Available for free at Connexions

91 This means that both electrons have their energy lowered, just as the single electron has its energy lower in H2 + . This means that, to separate the two H atoms, it takes more energy to raise the energies of these two electrons than to raise the energy of just one. So the bond energy of H2 is much greater than the bond energy of H2 + . From quantum mechanics, we can't add a third electron to this molecular orbital. We don't need the detail now to know what would happen if we added a third electron to an H2 molecule, but we can say that the energy of the third electron is not lowered. The strongest bond is formed by sharing just two electrons, not more and not less. And nally, why is the bond energy of H2 not double the bond energy of H2 + ? With two electrons in the same orbital, it would seem that we need double the energy of one electron to separate the atoms. Of course, this would assume that the two electrons are unaected by each other, with energies which do not depend on each other. This can't be true, since two electrons will repel each other by Coulomb's Law. In an atomic orbital, we saw that electron-electron repulsion raised the energy of both electrons sharing an orbital. The same is true in a molecular orbital. This means that the energy of each electron in H2 is not the same as the energy of the one electron in H2 + . In a discussion question, we consider the question of why this means that the bond energy in H2 is less than double the bond energy in H2 + .

9.5 Observation 3: Dipole moments in Diatomic Molecules We now have an understanding of what it means for two atoms to share an electron pair and why this results in a bond. In addition to H2 , this description works well in describing and understanding chemical bonds such as in F2 or Cl2 . More work is required to understand multiple bonds such as the double bond in O2 , but it turns out that the same principles apply. What about molecules where the atoms are not the same, e.g. HF? The Lewis model of these molecules still assumes a sharing of an electron pair. But we learned in our study of atomic structure that the properties of H atoms and F atoms are quite dierent. For example, the ionization energy of an F atom is larger than the ionization energy of an H atom. F atoms also have very strong electron anity. Do such dierent types of atoms share electrons? If so, does it matter that the properties of the two atoms are so dierent? We need observations to answer these questions. First, we can do as we did with H2 and examine the energies of the bonds between dierent atoms. The bond energy of HF is 568 kJ/mol, larger than the bond energy of H2 . By contrast, the bond energy of F2 is 154 kJ/mol, quite a bit weaker than the bonds in either HF or H2 . The bond energy of HCl is 432 kJ/mol, weaker than the HF bond. The energy of one O-H bond in H2 O is 463 kJ/mol. Clearly, the strength of bond depends on what types of atoms are bonded together. This suggests that the sharing of an electron pair depends on the properties of the atoms in the bond, including the sizes of the atoms and the charges on the nuclei. In fact, it would not be too surprising if dierent atoms did not share the electrons equally. To nd out if this is the case, we observe a property of molecules called the dipole moment. An electric dipole is simply a separation of a positive and a negative charge. The dipole moment, usually labeled µ, measures how strong the dipole is by taking the product of the amount of the charge times the separation of the charge. Just having a positive charge, let's say a proton, and a negative charge, let's say an electron, does not mean that there is a dipole moment. The hydrogen atom does not have a dipole moment because the electron is in constant rapid motion around the positive nucleus. As such, viewed from outside the atom, there is no positive end or negative end to the hydrogen atom, so there is no dipole moment. This is true of all atoms. It might seem that molecules could not have dipole moments for the same reason. The electrons are moving about the nuclei rapidly. In the H2 molecule, although there are positive and negative charges, there is no end of the molecule which looks more positive and no end which looks more negative. H2 does not have a dipole moment. What about a molecule like HF? If an H atom has no dipole moment and an F atom has no dipole moment, does HF have a dipole moment? Experimentally, molecular dipoles can be observed in a number of ways. If you put a molecule with a dipole moment in an electric eld, it will line up with the eld, with Available for free at Connexions

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CHAPTER 9. ENERGY AND POLARITY OF COVALENT CHEMICAL BONDS

the positive end of the dipole pointed towards the negative end of the eld. Dipoles can interact with each other as well, so that the negative end of one dipole will point towards the positive end of another dipole. It is also possible to measure the magnitude of the dipole moment. A number of these are given in Table 9.1: Dipole Moments of Specic Molecules for some simple molecules. We can see that, not only does HF have a dipole moment, but in fact it has the largest dipole moment of the molecules listed. What does it mean that HF has a dipole moment? An HF molecule must have a permanent negative end and a permanent positive end. Experimentation shows that the F end of the molecule has a net negative charge and the H end has a net positive charge.

Dipole Moments of Specic Molecules Molecule H2 O HF HCl HBr HI CO CO2 NH3 PH3 AsH3 CH4 NaCl

µ

(debye)

1.85 1.91 1.08 0.80 0.42 0.12 0 1.47 0.58 0.20 0 9.00

Table 9.1

How can this be so? The total number of electrons and protons in the molecule are evenly matched, of course, and the electrons are moving rapidly about the two nuclei just as in H2 . Perhaps the electrons are not moving uniformly around the H and the F nuclei. To observe this, we look at the molecular orbital for the shared electrons in the HF bond, shown in Figure 9.3.

Figure 9.3: Distribution of Electron Probability in HF

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93 We observe that the electrons in the molecule move with greater probability near the F atom. There is thus more electron charge around the F nucleus than the positive charge on the F nucleus. And the opposite is true for the H end of the molecule. Apparently, when an H atom and an F atom share electrons, they do not share them equally! This is an extremely important result, one of the most useful in all of Chemistry. Dierent atoms have dierent tendencies to draw electrons to themselves when sharing electrons in a covalent bond. The relative strength with which an atom draws electrons to itself in a bond is called electronegativity. A very electronegative atom will attract the shared electrons more strongly than a less electronegative atom. This will produce a negative charge near the more electronegative atom and a positive charge near the atom other. This will, in most cases, create a permanent dipole moment in the molecule. At this point, we don't know how much negative charge is near the more electronegative atom, but it probably is not the full charge on a single electron since the electrons are still mostly shared by the two atoms. As such, we label the negative charge by δ-, where δ is some number between 0 and 1. In most cases, we don't even need to know how large δ is. The positive end of the molecule is labeled by δ +, since whatever the negative charge on the negative end must equal the positive charge on the positive end. HF is a good starting example. The F atom is more electronegative than the H atom; hence, the HF molecule has a dipole moment, which makes HF what we call a polar molecule.

Figure 9.4

Of course, we next want to know why the F atom is more electronegative. We need more data to develop a model for electronegativity. Table 9.1: Dipole Moments of Specic Molecules listing the dipole moments of several molecules provides good data for comparison. Let's rst study the set of molecules HF, HCl, HBr, and HI. It is easy to see that the dipole moments of these molecules are in the order of HF > HCl > HBr > HI. To analyze this comparison, we need to remember that the dipole moment measures the product of the amount of charge separated times the distance by which the charges are separated. Since the dipole moment of HF is larger than that of HCl, let's compare these two. How do the properties of F atoms compare to those of Cl atoms? One thing we can say for sure: Cl is a larger atom than F. This means that the HCl bond (127 pm) is longer than the HF bond (92 pm). Since the dipole moment depends on the distance between the positive and negative charges, the comparison of bond lengths would perhaps lead us to predict that HCl would have a larger dipole moment than HF, But that's not what the data tell us. There is only one way to explain this. The dierence between HF and HCl must be in the amount of charge that is separated. It must be true that the negative charge on the F atom is greater than the negative charge on the Cl atom. This means that F attracts the shared electrons in the H-F bond more than the Cl atom attracts the shared electrons in the H-Cl bond. This means that F is more electronegative than Cl. Looking at the dipole moments of these four molecules and remembering that the atoms get larger in the order F < Cl < Br < I, it must be true that the electronegativities go in the order F > Cl > Br > I. In this group, the electronegativity is larger for smaller atoms. To see if this turns out to be generally true, we can examine other families in the periodic table. First, let's look at the dipole moments for H2 O and H2 S. H2 O has a dipole moment with the O being the negative end. This means that O is more electronegative than H. H2 S also has a dipole moment, but it is much smaller than that of H2 O, so S is less electronegative than O. Let's compare the dipole moments of NH3 and PH3 : we see the same trend. It is generally true that electronegativity is larger for smaller atoms. Available for free at Connexions

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We should also compare the electronegativities of elements in the same row of the Periodic Table. For example, the dipole moments increase in the order of NH3 < H2 O < HF, and PH3 < H2 S < HCl. From this, we can conclude that, in general, electronegativity increases with increasing atomic number within a single row of the Periodic Table. Electronegativity is an extremely useful concept in chemistry, but it is not a precisely dened physical property. In fact, there are several possible denitions of how to measure electronegativity each of which leads to slightly dierent values for each atom Although the exact values vary, the overall trends we observed are similar. Using one popular denition, Table 9.2: Electronegativity of Selected Atoms shows electronegativities for many atoms. Looking at these numbers, you should be able to see the trends we have developed from analyzing dipole moments of simple molecules.

Electronegativity of Selected Atoms Atom Electronegativity (X) H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca

2.1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.9 1.2 1.5 1.8 2.1 2.5 3.0 0.8 1.0

Table 9.2

With these observations in mind, we need to develop a model to understand why the electronegativity is larger for a larger atomic number in a single row, but is smaller for a larger atomic number in a single group. These trends seem to contradict each other. There must be a good physical explanation of these observations. To nd one, let's note that both trends point to F being the most electronegative element. F is as far to the right on the table as we can go and as far to the top of the table as we can go. (We might consider He or Ne to be more electronegative, but since there are no known molecules containing bonds with He or Available for free at Connexions

95 Ne, electronegativity has no meaning for these atoms.) What else do we know about F atoms? Thinking back on our understanding of atomic energy levels, we recall that F atoms have the highest ionization energy (except for He and Ne) and the highest electron anity of any of the elements. Electrons are clearly most strongly attracted to F atoms. We developed a model to explain this based on Coulomb's law. The F atom uniquely combines the largest core charge and the smallest distance of the valence electrons to the nucleus (measured as the average orbital distance). Perhaps these two factors are also responsible for F having the largest electronegativity. To nd out, we should examine other elements with high electronegativities. The highest electronegativities are all for elements with high ionization energies. Interesting examples include N, O, and Cl. Of these, O has the highest electronegativity, and this makes sense: it has a large core charge and a small shell radius. But if we compare N and O, N has the higher ionization energy. There must be more to electronegativity than just ionization energy. Another interesting comparison of N and O is that N has no electron anity whereas O has a strong electron anity, so electron anity must also be important in understanding electronegativity. This makes sense: an atom with a higher ionization energy is less likely to have its electrons drawn to another atom in a chemical bond, and an atom with a higher electron anity is more likely to draw electrons from another atom in a chemical bond. This produces a simple model to understand electronegativity. Atoms with higher ionization energies and higher electron anities have higher electronegativity. The reasons for high ionization energy and high electron anity are the same as the reasons for high electronegativity. On the basis of Coulomb's law, a larger core charge and a smaller shell radius generally give larger ionization energy, larger electron anity, and larger electronegativity. One way to dene electronegativity is simply as the average of the ionization energy and the electron anity. In fact, the values in Table 9.2: Electronegativity of Selected Atoms are just this average multiplied by a constant to give a simplied scale. Understanding these trends is extremely useful. Electronegativity is one of the most powerful concepts in chemistry for predicting chemical reactivity. For example, positive ends of molecules are often attracted to the negative ends of other molecules. Understanding where there may be a more negative charge in a molecule can then help us predict the location in a molecule where a reaction may take place or even predict whether a reaction is expected to occur or not. We will have many occasions to apply the concept of electronegativity, including in the next concept study.

9.6 Review and Discussion Questions 1. Why does an electron shared by two nuclei have a lower potential energy than an electron on a single atom? Why does an electron shared by two nuclei have a lower kinetic energy than an electron on a single atom? How does this sharing result in a stable molecule? How can this aect be measured experimentally? 2. The bond in an H2 molecule is almost twice as strong as the bond in the H2 + ion. Explain why the H2 bond is so much stronger. Why isn't the H2 bond exactly twice as strong as the H2 + bond? 3. The ionization energy of H2 is slightly less than the ionization energy of H2 + . But the bond energy of H2 is much larger than the bond energy of H2 + . Explain how these two facts are consistent with each other. 4. In this study, we referred to H2 as a stable molecule. But H2 gas can be explosively reactive, as a viewing of the Hindenberg disaster clearly reveals. In what sense is H2 stable? How can a stable molecule be a highly reactive molecule? 5. Explain why an atom with a high ionization energy is expected to have a high electronegativity. 6. Explain why an atom with a high electron anity is expected to have a high electronegativity. 7. Explain why S has a greater electronegativity than P but a smaller electronegativity than O. 8. N atoms have a high electronegativity. However, N atoms have no electron anity, meaning that N atoms do not attract electrons. Explain how and why these facts are not inconsistent.

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Chapter 10

Bonding in Metals and Metal-Non-Metal Salts 1

10.1 Introduction We have noted that the Lewis model of chemical bonding is very powerful in predicting structures, stability, and reactivity of molecules. But there is a glaring hole in our model that you may have noticed: the metal elements are missing. Additionally, the Lewis model only applies to a handful of atoms at a time, and we have not examined what happens in solids that have huge numbers of atoms bonded in vast networks. The Lewis model is based on the octet rule and the concept of a covalent bond as a sharing of an electron pair. These were developed based on the molecules formed by elements in Groups 4 to 8, and most specically, the group of elements we call the non-metals. This name clearly says that the properties of the non-metal elements are very dierent from the properties of metal elements. We will look at these dierences in this study. But even without analyzing those dierences, we can say immediately that the octet rule does not seem to apply to these elements. Remember that the octet rule says that the number of valence electrons plus the valence of the atom (the number of bonds the atom typically forms) commonly equals 8 for compounds formed by the non-metal elements. Rather than being the general rule for metals, this is very rarely true. This means that we need a new model for bonding in metals and in compounds that contain metal atoms. To develop this model, we will examine the specic properties of metallic elements, which dier signicantly from the non-metals. By considering these properties carefully, we will be able to build a model which accounts for these properties. Of course, to be more complete, we also need to consider compounds formed from combinations of metal atoms and non-metal atoms. These also have properties which dier greatly from either metals or nonmetals. Again, by looking closely at these properties, we will be able to build a model for metal-non-metal bonding, which is dierent from that in metal bonding. This means that we will develop models of two new types of bonding in addition to the one we have already developed for covalent bonding. It would be very helpful to nd a way to tie these three types of bonding together, to give a simple understanding of why the bonding is dierent for dierent types and combinations of atoms. In the last section of this study, we will create such a model based on our understanding of the chemical concept of electronegativity, developed in the previous concept study. 1 This content is available online at .

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10.2 Foundation In this study, we will assume that we know the essential components of the structure and properties of individual atoms. Each atom has an electronic conguration which determines its physical and chemical properties, including ionization energy, electron anity, atomic size, and electronegativity. Electron motion is described by orbitals, which give the probability for the electron in space around the nucleus. The energy of each electron is determined by a combination of its kinetic energy, its attraction to the nucleus, and its repulsion from other electrons in the atom. Our model considers the electron-electron repulsion as a shielding of the positive charge of the nucleus, resulting in an eective nuclear charge which is less than the actual nuclear charge, which we refer to as the core charge. By looking at the core charge experienced by an electron in an atom and at its distance from the nucleus, we can understand the ionization energy of that electron. We know and can account for the fact that the ionization energies are greatest for atoms near the right side of the periodic table with large core charges. And the ionization energies are greater for smaller atoms, where the valence electrons are closer to the nucleus. In a previous study, we developed the concept of electronegativity. An atom with a high electronegativity strongly attracts the shared electrons to itself in a covalent bond. The atom with the lower electronegativity in the bond more weakly attracts the shared electrons. The result is that the bond is polar, meaning that one end of the bond is negatively charged and the other end is positively charged. We will assume a previous understanding of the variations of electronegativity amongst the elements. Atoms to the right in the Periodic table have higher electronegativities than those to the left. And atoms in the earlier rows of the Periodic Table have higher electronegativities than those in the later rows. Electronegativity thus generally increases from left to right and down to up in the Periodic Table. These facts will be extremely useful in understanding how and why dierent types and combinations of atoms form dierent types of bonds.

10.3 Observation 1: Properties of Metals Historically, people have worked to locate, isolate, and purify metals because of their valuable properties. Most metals are both strong and malleable solids, meaning that they can be shaped, bent, pressed, attened, and so forth without cracking or breaking. This is a very useful property. Shelters, shields, tools, and armor can be made from solids provided that they can be bent to whatever shape is desired. Since they are not brittle, metals do not break on impact so they provide excellent protection as well as excellent materials for weapons. In the age of electricity, many metals became more valuable due to their conductivity. When a piece of metal is bridged across an electric potential, electrons ow from the negative electrode to the positive electrode, creating a current with obvious applications. By contrast, non-metals are rarely conductors and are more typically insulators. Adding to the applications of metals for electricity, metals are also ductile, meaning that they can be drawn into thin wires while maintaining strength. And not least, most metals are actually quite attractive, with shiny, smooth, colorful nishes. This gives metals intrinsic value in addition to their usefulness. It is not surprising that gold, silver, and copper have long been used for coins and jewelry, given their beauty and their resistance to oxidation. We can examine these properties of metals to try to understand how metal atoms are bonded together. The distinct properties of metals tells us that the bonding must be quite dierent from that in the covalent molecules of the non-metals we have been studying so far. These dierences must relate to the dierences in the properties of the individual atoms. So let's take a look at those properties. Perhaps the most important atomic property is, as we often have seen, the ionization energy of each metal atom. Figure 10.1 (Carbon Atoms in the Diamond Network Lattice) shows the rst ionization energy for each atom in the third and fourth rows of the Periodic Table, including both metals and non-metals. What trends do we see in these data? Two trends appear very clearly. One trend is that the ionization energies of metals are signicantly lower than the ionization energies of the non-metals. Another trend is that the ionization energies of the metals do not vary much from metal to metal. This is very dierent from the sharp increases we see in the non-metals as we move across the periodic table. Available for free at Connexions

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Ionization Energies of Dierent Elements Element Atomic Number (Z) First Ionization Energy (kJ/mol) Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

495.85 737.76 577.54 786.52 1011.82 999.60 1251.20 1520.58 418.81 589.83 633.09 658.82 650.92 652.87 717.28 762.47 760.41 737.13 745.49 906.41 578.85 762.18 944.46 940.97 1139.87 1350.77

Table 10.1

We have already discussed the rst trend in our study of atomic structure. We explained the lower ionization energies for metals compared to non-metals from the fact that the metals have relatively lower core charges. In our previous studies, we saw that the core charge increases for atoms as we increase the atomic number in a single row of the Periodic Table. Each row consists of elements with valence electrons in the same energy shell. The metals are more to the left in each row, meaning that they have smaller core charges for that row. This explains the relatively lower ionization energies. The lack of big variations in the ionization energies of metals is harder to understand. We have seen that the ionization energy of an atom is determined by the electron conguration. For the metal atoms, these electron congurations are a little trickier than for the non-metals. These are illustrated in Table 10.1: Available for free at Connexions

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Ionization Energies of Dierent Elements for the atoms in the fourth row of the Periodic Table, beginning with K. In each of these atoms, the 4s and 3d subshells have energies which are very close together. In K and Ca, the 4s orbital energy is lower, so the outermost electron or electrons in these two atoms are in the 4s orbital and there are no 3d electrons. For the transition metal atoms from V to Cu, these atoms have both 4s and 3d electrons and the number in each orbital depends very sensitively on exactly how many valence electrons there are and what the core charge is. As such, the actual electron congurations in Table 10.1: Ionization Energies of Dierent Elements would have been very hard to predict and we will treat them as data which we have observed.

Electron Congurations of Period 4 Elements K

Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

[Ar]4s1

[Ar]4s2 [Ar]3d1 4s2 [Ar]3d2 4s2 [Ar]3d3 4s2 [Ar]3d5 4s1 [Ar]3d5 4s2 [Ar]3d6 4s2 [Ar]3d7 4s2 [Ar]3d8 4s2 [Ar]3d10 4s1 [Ar]3d10 4s2 [Ar]3d10 4s2 4p1 [Ar]3d10 4s2 4p2 [Ar]3d10 4s2 4p3 [Ar]3d10 4s2 4p4 [Ar]3d10 4s2 4p5 [Ar]3d10 4s2 4p6 Table 10.2

A clear and surprising rule in the data in Table 10.1: Ionization Energies of Dierent Elements is that the outermost electron in each atom is always a 4s electron. As we increase the atomic number from V to Cu, there are more 3d electrons, but these are not the highest energy electrons in these atoms. Instead, these added 3d electrons increasingly shield the 4s electrons from the larger nuclear charge. The result is that there is not much increase in the core charge, so there is not much increase in the ionization energy, even with larger nuclear charge. This model explains the data in Figure 10.1 (Carbon Atoms in the Diamond Network Lattice). Although this analysis probably seems complicated, it helps us to understand the important observation that all of the metals have low ionization energies. How do these electron congurations determine the properties of metals? Or stated more specically, how do the electrons congurations aect the bonding of metal atoms to each other, and how does this bonding determine the properties of the metals? To nd out, let's look at each property and develop a model that accounts for it. First, think about the electrical conductivity of metals. When a relatively low Available for free at Connexions

101 electric potential is applied across a piece of metal, we observe a current, which is the movement of electrons through the metal from the negative to the positive end of the electric eld. The electrons in the metal respond fairly easily to that potential. For this to happen, at least some of the electrons in the metal must not be strongly attracted to their nuclei. Does this mean that they are somewhat loose in the metal? In fact, we have seen that this is true: the ionization energy of metal atoms is low. Perhaps the valence electrons are somewhat loose in the metal. A metal's conductivity tells us that when we have many metal atoms (let's say 1 mole, for example), there are electrons available to contribute to the current when an electric potential is applied. Thus, the valence electrons must not be localized to individual nuclei but rather are free to move about many nuclei. Second, let's think about the malleability and ductility of solid metals. These properties mean that the bonding of the metal atoms together is not aected much when the atoms are rearranged. It may be dicult to see on the macroscale, but bending a piece of metal or stretching into a thin wire requires major movement of atoms. And since bending the metal does not break it into pieces, the adjacent atoms must remain bonded together despite these large atomic movements. Apparently, the bonding electrons are not aected by this rearrangement of atoms. This is completely consistent with the idea we just discussed, that the electrons are free to move about many nuclei and are not just localized between two adjacent nuclei. When the atoms are rearranged by bending or stretching, the electrons are free to immediately rearrange as well, and the bonding is preserved. Our picture of a metal, based on these conclusions, is that the nuclei of the metal atoms are arranged in an array in the solid metal. The non-valence electrons in each metal, which are strongly attracted to each nucleus, remain localized near their own atoms. The valence electrons, though, are free to move about the positive centers of the nuclei and core electrons. Once you have this image in your head, you can see why chemists refer to this as the electron sea model of a metal. You should also be able to see how the properties of metals lead us to this electron sea image. What about the shininess of metals? To understand this, we need to know what causes light to shine o of a surface. From our previous studies, we learned that light (electromagnetic energy) can be absorbed by atoms causing electrons to move from a lower energy state to a higher one. Similarly, light can be emitted from an atom with an electron moving from a higher energy state to a lower one. According to Einstein's formula, the frequency of the light ν absorbed or emitted, when multiplied by a constant h, must match the energy dierence ∆E between the two electron states: ∆E=hν . Because there are so many electrons in the electron sea which are involved in the bonding of the metal atoms together, there are many, many electron energy levels, a huge number in fact. So there are a correspondingly huge number of energy dierences between these levels. This means that, when visible light hits the surface of a metal, the metal can easily absorb and reemit light of that frequency, reecting the light and making the surface appear to shine. Overall, we can see that the electron sea model of bonding of metal atoms together accounts for the properties of metals we have observed. It is worth thinking about how very dierent this model of bonding is from the covalent model of bonding in non-metals. We'll come back to this contrast in the last section of this study.

10.4 Observation 2: Properties of Salts There are many types of compounds formed by combining metals atoms and non-metal atoms. To simplify our discussion, we are going to focus on one specic type of compound called a salt. The common use of the term salt refers to one specic compound Sodium Chloride (NaCl), which is also a great example of the more general idea of a salt, so we'll start with it and then consider some more examples. What are the properties of NaCl? They are quite dierent than the properties of the metals we just discussed. First, NaCl is a solid crystal and it is not at all malleable. A crystal of NaCl, say rock salt, cannot be molded into whatever shape we choose. Rather, it is very brittle. Hit it with a hammer and, unlike a piece of metal, it shatters into tinier fragments of the crystal. Similarly, it is not ductile. It cannot be rolled or stretched into a wire or a thread. Second, solid NaCl is not an electrical conductor. Instead it Available for free at Connexions

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insulates against the movement of current even when an electric potential is applied. We can immediately conclude from these observations that the bonding model we developed for a metal is not going to work to describe bonding in NaCl. We'll have to start from scratch. There are other interesting properties of NaCl. One is that it dissolves easily in water, which most metals do not. And when dissolved in water, the resulting solution conducts electricity. Somehow then a current can pass through the salt solution, meaning that there are charged particles dissolved in the solution which carry the movement of charge. These charged particles turn out to be ions, Na+ and Cl- . Of course, this does not tell us whether there are ions in NaCl itself, since the interaction with the water molecules in the solution might change everything. Instead, we could try melting NaCl, so that we wind up with a liquid which is pure NaCl without any water. This takes a very high temperature, 808 ◦ C, indicating that there are strong forces at work in the solid NaCl crystal. When we melt NaCl, we nd that the resulting liquid does in fact conduct electricity. Liquid NaCl thus consists of ions, Na+ positive ions (cations) and Cl- negative ions (anions). As a result, we should expect that these same ions exist in the solid NaCl. How can we reconcile the existence of ions in solid NaCl with fact that it does not conduct an electric current? The answer is that a current is charge in motion. Thus, the simple existence of an ion is not enough to carry a current. The ion must also be able to move, as electrons do in a metal, or as Na+ and Cl- do when dissolved in water. The ions in the solid cannot move, at least not very far, as we have seen from the fact that NaCl is not malleable. In fact, the Na+ and Cl- ions are basically xed in place. From Coulomb's law, we know that opposite charges are strongly attracted to each other. We can conclude then that the bonding in NaCl is due to the attraction of Na+ cations to Cl- anions. Why are there ions in the solid? The solid crystal itself is not electrically charged, so it isn't clear why each Na atom has lost an electron and each Cl atom has gained an electron. Let's look again at the properties of these very dierent kinds of atoms. We know that Na has low ionization energy, but it isn't zero. It still does require a lot of energy to ionize the valence electron. We know that Cl has a much higher ionization energy. More importantly, we also know that Cl has a high electron anity, which means that a lot of energy is released when an electron is added to a Cl atom. Is the energy released when the electron is attached to the Cl atom enough to ionize the Na atom? To nd out, let's compare the ionization energy of Na to the electron anity of Cl: Na → Na+ + e- Ionization energy = 496 kJ/mol Cl + e- → Cl- Electron anity = -349 kJ/mol Our answer is no. Taking an electron from a Na atom and giving it to a Cl atom costs a good amount of energy in total. This seems to suggest that the electron should not leave the Na atom and join the Cl atom, so NaCl shouldn't form ions and therefore shouldn't form a stable compound. We haven't considered one key factor, however. The energy comparison above leads to the formation of independent positive and negative ions which don't interact with each other after the reaction is complete. But in reality, the Na+ and Cl- ions are very close to one another and attracted to one another. Coulomb's law tells us that this signicantly lowers the energy. And there is even more to consider. A crystal of NaCl does not consist of a single Na+ and a single Cl- . Instead, it is an entire array of many positive and negative ions. Each positive ion is surrounded by several negative ions. And each negative ion is surrounded by the same number of positive ions. (It turns out that number is 6.) Coulomb's law tells us that we get a huge lowering of energy from having all these opposite charges adjacent to one another. This energy is called the lattice energy and it is very large, -787 kJ/mol. This is much more than the energy decit for ionizing both atoms, and accounts easily for the bonding in NaCl. The bonding in NaCl is thus dierent than the covalent bonding in, say, HF or the metallic bonding in, say, Cu metal. For obvious reasons, we refer to this type of bonding as ionic bonding. Before concluding that ionic bonding is responsible for the stability of NaCl, we need to ask about the other primary property of NaCl mentioned above. Specically, NaCl is brittle and not malleable. This is quite dierent from the property of a metal. In a metal, we could rearrange the atoms, for example by bending or by deforming with a hammer, and the atoms remain strongly bonded. But we cannot bend NaCl crystals, and if we hit them with a hammer, the bonding is destroyed as the crystal shatters. We simply Available for free at Connexions

103 cannot rearrange the atoms. It is clear that the bonding in NaCl depends very much on the arrangement of the atoms. If we think about our ionic bonding model, this makes perfect sense. For the ionic bonding to work, the negative ions must remain surrounded by the positive ions and vice versa. Any attempt to rearrange these ions will result in positive ions adjacent to positive ions and negative ions next to negative ions. This will create strong repulsions, and the solid will fall apart. Ionic bonding thus accounts for the brittleness of NaCl. So far, we've only looked at ionic bonding in NaCl as an example, but since we've seen that dierent covalent bonds have dierent energies, perhaps dierent ionic bonds have dierent energies. We compare dierent salts to see if there are dierent lattice energies in the ionic bonds. Table 10.2: Electron Congurations of Period 4 Elements shows a set of lattice energies for salts formed from alkali metals (Li, Na, K, Rb) and halogens (F, Cl, Br, I). There are some clear trends in these data. The largest lattice energy corresponds to the combination of the two smallest ions, Li+ and F- . The lattice energy decreases when either or both of the ions are larger, with the smallest being for RbI, consisting of the two largest ions.

Lattice Energies for Alkali Halides (kJ/mol) Li + Na + K+ Rb +

F  Cl  Br  I  1036 923 821 785

853 787 715 689

807 747 682 660

757 704 649 630

Table 10.3

Why would size be a determining factor in the lattice energy? We should recall that the lattice energy follows Coulomb's law. So, the closer the charges are to one another, the stronger is the interaction. Smaller ions can be closer together than larger ions. So the lattice energy is largest for the smallest ions. Of course, Coulomb's law also involves the number of the charges. In all of the compounds in Table 10.2: Electron Congurations of Period 4 Elements, the ions have a single +1 or -1 charge. We should look at compounds which contain doubly-charged ions. For common ions with +2 charges, we can look at the alkali earth metals In Table 10.3: Lattice Energies for Alkali Halides, we can easily see that the lattice energies for salts of these ions are much larger than for the alkali metal ions. One nal comparison would be a doubly-charged negative ion like O2- . Again, the lattice energies involving single positive charges with O2- are larger, and the lattice energy is even larger still when both ions are doubly charges, as in MgO.

Lattice Energies for Alkaline Earth Halides and Oxides (kJ/mol) (kJ/mol) Mg 2+ Ca 2+ Sr 2+ Ba 2+

F-

2936 2608 2475 2330

Cl - Br - I 2496 2226 2127 2028

2397 2131 2039 1948

2289 2039 1940 1845

O 23923 3517 3312 3120

Table 10.4

We can conclude that compounds of metals and non-metals are typically formed by ionic bonding, and the strength of this bonding can be clearly understood using Coulomb's law. Available for free at Connexions

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10.5 Observation 3: Properties and Bonding in Solid Carbon In the rst two observations of this study, we considered bonding in solids of two types, metals and salts. These are just two of the many types of solids, and not all solids are formed by either ionic bonding or metallic bonding. Far from it. We cannot look at every type of solid in this study, but it is worth considering one specic example which forms an interesting contrast to metals and salts. This example is diamond, one of several forms of pure solid carbon. (The other primary forms are graphene and the set of materials called fullerenes. We will postpone study of those materials for later.) As always, we should begin with experimental observations to guide our understanding of diamond. What are its primary properties? It is a very hard solid, generally regarded as the hardest solid available in bulk. It is not malleable and would not be considered brittle like NaCl. It can be cleaved only with signicant force. It has a very, very high melting point, over 3500 ◦ C, and as an interesting note, it is the most thermally conducting material we know, meaning that it transfers heat better than any other substance. However, it does not conduct electricity. What can we infer about the bonding in diamond from these properties? Since it is not brittle, we do not expect ionic bonding in diamond. This makes sense, since all of the atoms are carbon. But since it does not conduct electricity, we do not expect that the electrons are delocalized over the entire crystal, as they are in a metal. The bonding electrons must be more localized to individual nuclei. Since diamond is very hard and not malleable, the bonding must depend on the specic arrangement of the atoms, since, unlike a metal, it is very dicult to rearrange these atoms. Diamond won't bend! So diamond has neither metallic bonding nor ionic bonding. But this is not a surprise, since we know already that carbon forms covalent bonds. We can recall that carbon atoms have a valence of 4 and have 4 valence electrons which they commonly share with other carbon atoms or with non-metal atoms to form covalent bonds. Since all of our observations suggest that the bonding electrons in diamond are localized, we can imagine that all of the bonding in diamond is covalent. That makes sense, based on our knowledge of the ways that carbon atoms bond with dierent elements. Let's pick one carbon atom to start with. What if that carbon atom is bonded to four other carbon atoms, satisfying the octet rule for covalent bonding? And what if each of those four carbon atoms is, in turn, bonded to three additional carbon atoms. And those twelve carbon atoms are, again in turn, bonded to three additional carbon atoms. We can build an entire network of carbon atoms this way, as is illustrated in Figure 10.1 (Carbon Atoms in the Diamond Network Lattice) where the C atoms are shown with bonds connecting them.

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Carbon Atoms in the Diamond Network Lattice

Figure 10.1

Looking closely at this, it is clear that each carbon atom has a complete octet, satisfying its valence of 4. Thinking about this more carefully, it is also clear that this network does not have to end. We could continue building this network by adding more and more carbon atoms, each bonded to four other carbon atoms, building a huge molecule large enough to hold. Or to wear on a ring! Does the bonding model in Figure 10.1 (Carbon Atoms in the Diamond Network Lattice) account for the properties of diamond? The electrons are all localized, so we wouldn't expect diamond to conduct electricity. The atoms are very precisely arranged in the network and cannot be moved relative to one another without breaking some of the covalent bonds, so diamond is very hard and non-malleable. We call the bonding in diamond network covalent, since the bonding is all covalent and creates a network of carbon atoms. There is an important unanswered question: why do the carbon atoms sit in the particular geometry as they do in Figure 10.1 (Carbon Atoms in the Diamond Network Lattice)? The answer is not obvious right now, and the question is the subject of the next concept study. For now, we'll simply note that a carbon atom with four single bonds will arrange those bonds in the shape of a tetrahedron. When all of the carbon atoms are networked together with this geometry, we get the network in Figure 10.1 (Carbon Atoms in the Diamond Network Lattice).

10.6 A Model for Predicting the type of Bonding: Electronegativity We have now seen three types of bonding in solids. In metallic bonding, the bonding valence electrons are delocalized in an electron sea, allowing current to ow and permitting distortions of the arrangements of the atoms. In ionic bonding, adjacent positive metal ions and negative non-metal ions are strongly attracted to each other in an array which places positive ions next to negative ions and vice versa. This creates a hard, Available for free at Connexions

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brittle solid, not permitting rearrangement of the ions and not allowing electron ow in an electric current. In a solid covalent network like diamond, the bonding is the more familiar covalent sharing of an electron pair so that each non-metal atom in the network satises its valence in agreement with the octet rule. This makes a very hard solid which is neither brittle nor malleable, and this does not allow movement of electrons in a current. It would be nice to add to our bonding model a way to understand or even to predict which of these types of bonding is expected for a particular solid. This would allow us to understand or predict the properties of the solid from the properties of the atoms which make it up. How shall we begin? The most consistent trend we have seen is that bonding appears quite dierent for metals, non-metals, and combinations of metals and non-metals. At least from what we have observed so far, metal atoms bond to metal atoms with metallic bonding (hence the name!), metal atoms bond to non-metal atoms with ionic bonding, and non-metal atoms bond to non-metal atoms with covalent bonding. This suggests that we look at the dierences between metal atoms and non-metal atoms. From our previous concept study, we know one major dierence: the electronegativity of non-metals is quite high, whereas the electronegativity of metals is typically much lower. Let's break this down in terms of the three types of bonding. The easiest case is ionic bonding. In this case, we have combined a metal with a non-metal, like Na with Cl, so we have combined atoms with high electronegativity with atoms with low electronegativity. Apparently, atoms with these properties tend to attract each other with ionic bonding. This makes sense: with very dierent electronegativities, the atoms are not likely to share bonding electrons. It is likely that the very electronegative atoms will form negative ions and the weakly electronegative atoms with form positive ions, and the oppositely charged ions will attract each other. Thus, our model can be that, when a compound contains atoms with very dierent electronegativities, the compound is likely to be ionic bonded and have the properties of an ionic solid. By process of elimination, the remaining types of bonding, metallic and covalent, must involve atoms with similar electronegativities. But something must distinguish these two in a way that we can predict. Metallic bonding is expected when all the atoms are metals and therefore have low electronegativity. Covalent bonding is expected when all the atoms are non-metals and therefore have relatively high electronegativity. Here is the general summary for our model: • When the bonded atoms in a compound all have low electronegativity, we should predict metallic bonding and the compound should be a solid with metallic properties. • When the bonded atoms in a compound have very dierent electronegativities, we should predict ionic bonding and the compound should be a brittle, non-conducting solid. • When the bonded atoms in a compound all have high electronegativity, we should predict covalent bonding. In the last case, there are many types of solids possible, and the properties of a covalent compound depend very much on the types of solid which is formed. To illustrate, diamond and ammonia NH3 are both covalent compounds, but the properties of these two compounds could hardly be more dierent. We'll need a more extensive model to predict what type of covalent compound will form. This model leads to a simple picture for a reasonable prediction of the type of bonding. We need to consider both the dierences in electronegativity between the atoms in a compound as well as the actual magnitudes of the electronegativities, either high or low. This means that, at least for binary compounds (those involving only 2 elements), we can create a chart showing both the magnitude of the electronegativities of the atoms (taken as an average of the two electronegativities) and the dierence between the two electronegativities. We wind up with a chart that looks like a triangle, as in Figure 10.2. In fact, this is called a bond type triangle. A few compounds are shown on the triangle to illustrate how this model can be used successfully to predict the type of solid from the atoms involved.

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Figure 10.2

10.7 Discussion Questions 1. Explain why the low ionization energy of a metal atom is important to the bonding in the metal. 2. The four metals Sc, Ti, V, and Cr have increasing nuclear charge in the order listed but have only small dierences in ionization energy. On the basis of the electron congurations of these elements, explain this unexpected lack of variation. 3. It is often argued that alkali metals form ionic bonds with halogens (e.g. NaCl) because the Na atom can form an octet of electrons by losing it valence electron, thus lowering its energy. Using experimental data, demonstrate that this is an incorrect model for explaining the formation of an ionic bond. Available for free at Connexions

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4. From the data in Table 10.2: Electron Congurations of Period 4 Elements and Table 10.3: Lattice Energies for Alkali Halides, identify three trends in the variation of lattice energies amongst these compounds, and explain these variations on the basis of Coulomb's law. 5. When a metal can be shaped into a spring, the metal can be stretched but will return to it original shape when released. Explain this behavior in terms of the bonding model of a metal. Suggest a reason why a salt cannot be used to make a spring. 6. Looking at the carbon atom network in diamond illustrated in Figure 10.1 (Carbon Atoms in the Diamond Network Lattice), we could imagine that the bonding electrons are delocalized over the entire network, as in our electron sea model of a metal. Provide and explain experimental data which demonstrate that the electron sea model does not apply to solid carbon. 7. Why is it necessary to consider both electronegativity dierences between bonded atoms and the average electronegativity of bonded atoms when analyzing the type of bond which is formed? 8. Why is the bond type diagram in Figure 10.2 a triangle? That is, why is it not possible to observe compounds over the entire range of average electronegativity and electronegativity dierence?

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Chapter 11

Molecular Geometry and Electron Domain Theory 1

11.1 Foundation

We begin by assuming a Lewis structure model for chemical bonding based on valence shell electron pair sharing and the octet rule. We thus assume the nuclear structure of the atom, and we further assume the existence of a valence shell of electrons in each atom which dominates the chemical behavior of that atom. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. A number of atoms, including C, N, O, P, and S, can form double or triple bonds as needed to complete an octet. We know that double bonds are generally stronger and have shorter lengths than single bonds, and triple bonds are stronger and shorter than double bonds.

11.2 Goals We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. The relationship between bonding, structure, and properties is comparatively simple in diatomic molecules, which contain two atoms only, e.g. HCl or O2 . A polyatomic molecule contains more than two atoms. An example of the complexities which arise with polyatomic molecules is molecular geometry: how are the atoms in the molecule arranged with respect to one another? In a diatomic molecule, only a single molecular geometry is possible since the two atoms must lie on a line. However, with a triatomic molecule (three atoms), there are two possible geometries: the atoms may lie on a line, producing a linear molecule, or not, producing a bent molecule. In molecules with more than three atoms, there are many more possible geometries. What geometries are actually observed? What determines which geometry will be observed in a particular molecule? We seek a model which allows us to understand the observed geometries of molecules and thus to predict these geometries. Once we have developed an understanding of the relationship between molecular structure and chemical bonding, we can attempt an understanding of the relationship of he structure and bonding in a polyatomic molecule to the physical and chemical properties we observe for those molecules. 1 This content is available online at .

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CHAPTER 11. MOLECULAR GEOMETRY AND ELECTRON DOMAIN THEORY 11.3 Observation 1: Geometries of molecules 110

The geometry of a molecule includes a description of the arrangements of the atoms in the molecule. At a simple level, the molecular structure tells us which atoms are bonded to which. At a more detailed level, the geometry includes the lengths of all of these bonds, that is, the distances between the atoms which are bonded together, and the angles between pairs of bonds. For example, we nd that in water, H2 O, the two hydrogens are bonded to the oxygen and each O-H bond length is 95.72 pm (where 1 pm = 10-12 m). Furthermore, H2 O is a bent molecule, with the H-O-H angle equal to 104.5 ◦ . (The measurement of these geometric properties is dicult, involving the measurement of the frequencies at which the molecule rotates in the gas phase. In molecules in crystalline form, the geometry of the molecule is revealed by irradiating the crystal with x-rays and analyzing the patterns formed as the x-rays diract o of the crystal.) Not all triatomic molecules are bent, however. As a common example, CO2 is a linear molecule. Larger polyatomics can have a variety of shapes, as illustrated in Figure 11.1 (Molecular Structures). Ammonia, NH3 , is a pyramid-shaped molecule, with the hydrogens in an equilateral triangle, the nitrogen above the plane of this triangle, and a H-N-H angle equal to 107 ◦ . The geometry of CH4 is that of a tetrahedron, with all H-C-H angles equal to 109.5 ◦ . (See also Figure 11.2(a).) Ethane, C2 H6 , has a geometry related to that of methane. The two carbons are bonded together, and each is bonded to three hydrogens. Each H-C-H angle is 109.5 ◦ and each H-C-C angle is 109.5 ◦ . By contrast, in ethene, C2 H4 , each H-C-H bond angle is 116.6 ◦ and each H-C-C bond angle is 121.7 ◦ . All six atoms of ethene lie in the same plane. Thus, ethene and ethane have very dierent geometries, despite the similarities in their molecular formulae.

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Molecular Structures

Figure 11.1

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We begin our analysis of these geometries by noting that, in the molecules listed above which do not contain double or triple bonds (H2 O, NH3 , CH4 and C2 H6 ), the bond angles are very similar, each equal to or very close to the tetrahedral angle 109.5 ◦ . To account for the observed angle, we begin with our valence shell electron pair sharing model, and we note that, in the Lewis structures of these molecules, the central atom in each bond angle of these molecules contains four pairs of valence shell electrons. For methane and ethane, these four electron pairs are all shared with adjacent bonded atoms, whereas in ammonia or water, one or two (respectively) of the electron pairs are not shared with any other atom. These unshared electron pairs are called lone pairs . Notice that, in the two molecules with no lone pairs, all bond angles are exactly equal to the tetrahedral angle, whereas the bond angles are only close in the molecules with lone pairs. One way to understand this result is based on the mutual repulsion of the negative charges on the valence shell electrons. Although the two electrons in each bonding pair must remain relatively close together in order to form the bond, dierent pairs of electrons should arrange themselves in such a way that the distances between the pairs are as large as possible. Focusing for the moment on methane, the four pairs of electrons must be equivalent to one another, since the four C-H bonds are equivalent, so we can assume that the electron pairs are all the same distance from the central carbon atom. How can we position four electron pairs at a xed distance from the central atom but as far apart from one another as possible? A little reection reveals that this question is equivalent to asking how to place four points on the surface of a sphere spread out from each other as far apart as possible. A bit of experimentation reveals that these four points must sit at the corners of a tetrahedron, an equilateral triangular pyramid, as may be seen in Figure 11.2(b). If the carbon atom is at the center of this tetrahedron and the four electron pairs at placed at the corners, then the hydrogen atoms also form a tetrahedron about the carbon. This is, as illustrated in Figure 11.2(a), the correct geometry of a methane molecule. The angle formed by any two corners of a tetrahedron and the central atom is 109.5 ◦ , exactly in agreement with the observed angle in methane. This model also works well in predicting the bond angles in ethane.

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Tetrahedral Structure of Methane

(a)

(b) Figure 11.2:

atom.

(a) The dotted lines illustrate that the hydrogens form a tetrahedron about the carbon

(b) The same tetrahedron is formed by placing four points on a sphere as far apart from one

another as possible.

We conclude that molecular geometry is determined by minimizing the mutual repulsion of the valence shell electron pairs. As such, this model of molecular geometry is often referred to as the valence shell electron pair repulsion (VSEPR) theory . For reasons that will become clear, extension of this model implies that a better name is the Electron Domain (ED) Theory . This model also accounts, at least approximately, for the bond angles of H2 O and Nh3 . These molecules are clearly not tetrahedral, like CH4 , since neither contains the requisite ve atoms to form the tetrahedron. However, each molecule does contain a central atom surrounded by four pairs of valence shell electrons. We expect from our Electron Domain model that those four pairs should be arrayed in a tetrahedron, without regard to whether they are bonding or lone-pair electrons. Then attaching the hydrogens (two for oxygen, Available for free at Connexions

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three for nitrogen) produces a prediction of bond angles of 109.5 ◦ , very close indeed to the observed angles of 104.5 ◦ in H2 O and 107 ◦ in NH3 . Note, however, that we do not describe the geometries of H2 O and NH3 as "tetrahedral," since the atoms of the molecules do not form tetrahedrons, even if the valence shell electron pairs do. (It is worth noting that these angles are not exactly equal to 109.5 ◦ , as in methane. These deviations will be discussed later (Section 11.5: Observation 3: Distortions from Expected Geometries).) We have developed the Electron Domain model to this point only for geometries of molecules with four pairs of valence shell electrons. However, there are a great variety of molecules in which atoms from Period 3 and beyond can have more than an octet of valence electrons. We consider two such molecules illustrated in Figure 11.3 (More Molecular Structures).

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More Molecular Structures

Figure 11.3

First, PCl5 is a stable gaseous compound in which the ve chlorine atoms are each bonded to the phosphorous atom. Experiments reveal that the geometry of PCl5 is that of a trigonal bipyramid: three of the chlorine atoms form an equilateral triangle with the P atom in the center, and the other two chlorine atoms are on top of and below the P atom. Thus there must be 10 valence shell electrons around the phosphorous atom. Hence, phosphorous exhibits what is called an expanded valence in PCl5 . Applying our Electron Domain model, we expect the ve valence shell electron pairs to spread out optimally to minimize their repulsions. The required geometry can again be found by trying to place ve points on the surface of a sphere with maximum distances amongst these points. A little experimentation reveals that this can be achieved by placing the ve points to form a trigonal bipyramid. Hence, Electron Domain theory Available for free at Connexions

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accounts for the geometry of PCl5 . Second, SF6 is a fairly unreactive gaseous compound in which all six uorine atoms are bonded to the central sulfur atom. Again, it is clear that the octet rule is violated by the sulfur atom, which must therefore have an expanded valence. The observed geometry of SF6 , as shown in Figure 11.3 (More Molecular Structures), is highly symmetric: all bond lengths are identical and all bond angles are 90 ◦ . The F atoms form an octahedron about the central S atom: four of the F atoms form a square with the S atom at the center, and the other two F atoms are above and below the S atom. To apply our Electron Domain model to understand this geometry, we must place six points, representing the six electron pairs about the central S atom, on the surface of a sphere with maximum distances between the points. The requisite geometry is found, in fact, to be that of an octahedron, in agreement with the observed geometry. As an example of a molecule with an atom with less than an octet of valence shell electrons, we consider boron trichloride, BCl3 . The geometry of BCl3 is also given in Figure 11.3 (More Molecular Structures): it is trigonal planar , with all four atoms lying in the same plane, and all Cl-B-Cl bond angles equal to 120 ◦ . The three Cl atoms form an equilateral triangle. The Boron atom has only three pairs of valence shell electrons in BCl3 . In applying Electron Domain theory to understand this geometry, we must place three points on the surface of a sphere with maximum distance between points. We nd that the three points form an equilateral triangle in a plane with the center of the sphere, so Electron Domain is again in accord with the observed geometry. We conclude from these predictions and observations that the Electron Domain model is a reasonably accurate way to understand molecular geometries, even in molecules which violate the octet rule.

11.4 Observation 2: Molecules with Double or Triple Bonds In each of the molecules considered up to this point, the electron pairs are either in single bonds or in lone pairs. In current form, the Electron Domain model does not account for the observed geometry of C2 H4 , in which each H-C-H bond angle is 116.6 ◦ and each H-C-C bond angle is 121.7 ◦ and all six atoms lie in the same plane. Each carbon atom in this molecule is surrounded by four pairs of electrons, all of which are involved in bonding, i.e. there are no lone pairs. However, the arrangement of these electron pairs, and thus the bonded atoms, about each carbon is not even approximately tetrahedral. Rather, the H-C-H and H-C-C bond angles are much closer to 120 ◦ , the angle which would be expected if three electron pairs were separated in the optimal arrangement, as just discussed for BCl3 . This observed geometry can be understood by re-examining the Lewis structure. Recall that, although there are four electron pairs about each carbon atom, two of these pairs form a double bond between the carbon atoms. It is tempting to assume that these four electron pairs are forced apart to form a tetrahedron as in previous molecules. However, if this were this case, the two pairs involved in the double bond would be separated by an angle of 109.5 ◦ which would make it impossible for both pairs to be localized between the carbon atoms. To preserve the double bond, we must assume that the two electron pairs in the double bond remain in the same vicinity. Given this assumption, separating the three independent groups of electron pairs about a carbon atom produces an expectation that all three pairs should lie in the same plane as the carbon atom, separated by 120 ◦ angles. This agrees very closely with the observed bond angles. We conclude that the our model can be extended to understanding the geometries of molecules with double (or triple) bonds by treating the multiple bond as two electron pairs conned to a single domain. It is for this reason that we refer to the model as Electron Domain theory. Applied in this form, Electron Domain theory can help us understand the linear geometry of CO2 . Again, there are four electron pairs in the valence shell of the carbon atom, but these are grouped into only two domains of two electron pairs each, corresponding to the two C=O double bonds. Minimizing the repulsion between these two domains forces the oxygen atoms to directly opposite sides of the carbon, producing a linear molecule. Similar reasoning using Electron Domain theory as applied to triple bonds correctly predicts that acetylene, HCCH, is a linear molecule. If the electron pairs in the triple bond are treated as a single domain, then each carbon atom has only two domains each. Forcing these domains to opposite sides from one another accurately predicts 180 ◦ H-C-C bond angles. Available for free at Connexions

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11.5 Observation 3: Distortions from Expected Geometries It is interesting to note that some molecular geometries (CH4 , CO2 , HCCH) are exactly predicted by the Electron Domain model, whereas in other molecules, the model predictions are only approximately correct. For examples, the observed angles in ammonia and water each dier slightly from the tetrahedral angle. Here again, there are four pairs of valence shell electrons about the central atoms. As such, it is reasonable to conclude that the bond angles are determined by the mutual repulsion of these electron pairs, and are thus expected to be 109.5 ◦ , which is close but not exact. One clue as to a possible reason for the discrepancy is that the bond angles in ammonia and water are both less than 109.5 ◦ . Another is that both ammonia and water molecules have lone pair electrons, whereas there are no lone pairs in a methane molecule, for which the Electron Domain prediction is exact. Moreover, the bond angle in water, with two lone pairs, is less than the bond angles in ammonia, with a single lone pair. We can straightforwardly conclude from these observations that the lone pairs of electrons must produce a greater repulsive eect than do the bonded pairs. Thus, in ammonia, the three bonded pairs of electrons are forced together slightly compared to those in methane, due to the greater repulsive eect of the lone pair. Likewise, in water, the two bonded pairs of electrons are even further forced together by the two lone pairs of electrons. This model accounts for the comparative bond angles observed experimentally in these molecules. The valence shell electron pairs repel one another, establishing the geometry in which the energy of their interaction is minimized. Lone pair electrons apparently generate a greater repulsion, thus slightly reducing the angles between the bonded pairs of electrons. Although this model accounts for the observed geometries, why should lone pair electrons generate a greater repulsive eect? We must guess at a qualitative answer to this question, since we have no description at this point for where the valence shell electron pairs actually are or what it means to share an electron pair. We can assume, however, that a pair of electrons shared by two atoms must be located somewhere between the two nuclei, otherwise our concept of "sharing" is quite meaningless. Therefore, the powerful tendency of the two electrons in the pair to repel one another must be signicantly oset by the localization of these electrons between the two nuclei which share them. By contrast, a lone pair of electrons need not be so localized, since there is no second nucleus to draw them into the same vicinity. Thus more free to move about the central atom, these lone pair electrons must have a more signicant repulsive eect on the other pairs of electrons. These ideas can be extended by more closely examining the geometry of ethene, C2 H4 . Recall that each H-C-H bond angle is 116.6 ◦ and each H-C-C bond angle is 121.7 ◦ , whereas the Electron Domain theory prediction is for bond angles exactly equal to 120 ◦ . We can understand why the H-C-H bond angle is slightly less than 120 ◦ by assuming that the two pairs of electrons in the C=C double bond produce a greater repulsive eect than do either of the single pairs of electrons in the C-H single bonds. The result of this greater repulsion is a slight "pinching" of the H-C-H bond angle to less than 120 ◦ . The concept that lone pair electrons produce a greater repulsive eect than do bonded pairs can be used to understand other interesting molecular geometries. Sulfur tetrauoride, SF4 , is a particularly interesting example, shown in Figure 11.4 (Molecular Structure of SF4 ).

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Molecular Structure of SF4

Figure 11.4

Note that two of the uorines form close to a straight line with the central sulfur atom, but the other two are approximately perpendicular to the rst two and at an angle of 101.5 ◦ to each other. Viewed sideways, this structure looks something like a seesaw. To account for this structure, we rst prepare a Lewis structure. We nd that each uorine atom is singly bonded to the sulfur atom, and that there is a lone pair of electrons on the sulfur. Thus, with ve electron pairs around the central atom, we expect the electrons to arrange themselves in a trigonal bipyramid, similar to the arrangement in PCl5 in Figure 11.3 (More Molecular Structures). In this case, however, the uorine atoms and the lone pair could be arranged in two dierent ways with two dierent resultant molecular structures. The lone pair can either go on the axis of the trigonal bipyramid (i.e. above the sulfur) or on the equator of the bipyramid (i.e. beside the sulfur). The actual molecular structure in Figure 11.4 (Molecular Structure of SF4 ) shows clearly that the lone pair goes on the equatorial position. This can be understood if we assume that the lone pair produces a greater repulsive eect than do the bonded pairs. With this assumption, we can deduce that the lone pair should be placed in the trigonal bipyramidal arrangement as far as possible from the bonded pairs. The equatorial position does a better job of this, since only two bonding pairs of electrons are at approximately 90 ◦ angles from the lone pair in this position. By contrast, a lone pair in the axial position is approximately 90 ◦ away from three bonding pairs. Therefore, our Electron Domain model assumptions are consistent with the observed geometry of SF4 . Note that these assumptions also correctly predict the observed distortions away from the 180 ◦ and 120 ◦ angles which would be predicted by a trigonal bipyramidal arrangement of the ve electron pairs.

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11.6 Review and Discussion Questions 1. Using a styrofoam or rubber ball, prove to yourself that a tetrahedral arrangement provides the maximum separation of four points on the surface of the ball. Repeat this argument to nd the expected arrangements for two, three, ve, and six points on the surface of the ball. 2. Explain why arranging points on the surface of a sphere can be considered equivalent to arranging electron pairs about a central atom. 3. The valence shell electron pairs about the central atom in each of the molecules H2 O, NH3 , and CH4 are arranged approximately in a tetrahedron. However, only CH4 is considered a tetrahedral molecule. Explain why these statements are not inconsistent. 4. Explain how a comparison of the geometries of H2 O and CH4 leads to a conclusion that lone pair electrons produce a greater repulsive eect than do bonded pairs of electrons. Give a physical reason why this might be expected. 5. Explain why the octet of electrons about each carbon atom in ethene, C2 H4 , are not arranged even approximately in a tetrahedron. 6. Assess the accuracy of the following reasoning and conclusions:

A trigonal bipyramid forms when there are ve electron domains. If one ED is a lone pair, then the lone pair takes an equatorial position and the molecule has a seesaw geometry. If two EDs are lone pairs, we have to decide among the following options: both axial, both equatorial, or one axial and one equatorial. By placing both lone pairs in the axial positions, the lone pairs are as far apart as possible, so the trigonal planar structure is favored.

7. Assess the accuracy of the following reasoning and conclusions:

The Cl-X-Cl bond angles in the two molecules (Figure 11.5) are identical, because the bond angle is determined by the repulsion of the two Cl atoms, which is identical in the two molecules.

Figure 11.5

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Chapter 12

Measuring Energy Changes in Chemical Reactions 1

12.1 Introduction Energy is a central theme of the study of Chemistry. The most common chemical reactions are carried out entirely for their production of energy. The most common sources of energy for use by humans are all chemical reactions. And the source of energy in the human body is entirely from chemical reactions. The industries of production, transportation, storage, and conversion of energy sources are overwhelmingly chemically based. To this point in our studies, we have discussed energy extensively but only to help us understand the stability of atoms, the electronic structure of atoms, the stability of chemical bonds, the geometry of molecules, the bonding in metals and salts, and so forth. We have not yet studied the energy changes which accompany chemical reactions. This study and the next mark a signicant transition in our studies. Rather than focus entirely on the structure of atoms and molecules, we will now consider observations of chemical properties on the macroscopic level. One of the major goals of developing chemical models is to relate the structures of individual atoms and molecules to the properties we observe for substances and reactions in bulk amounts. This might seem an insurmountable task. Since a bulk sample of a substance may contain literally trillions of trillions of particles, relating the properties of those particles to the properties of the bulk seems to require an incomprehensible amount of information. In this study and those that follow, we begin that process. First, we make detailed observations about the amounts of energy which are released or absorbed during chemical reactions and develop a method for measuring reaction energies for bulk reactions. Then, in the next study, we will relate the energies of chemical bonds to the energies of chemical reactions. But rst, we must relate the energies of chemical reactions to things we can measure directly and easily.

12.2 Introduction To make any progress with energy in Chemistry, we must assume some basic principles about energy from Physics. Energy is the capacity to do work, where work is the application of a force over a distance. We can therefore tell whether an object possesses energy by determining whether it has the capacity to accelerate another object. Keep in mind that this is capacity to do work. It is not necessary for an object to actually do work to have energy. We often speak two broad types of energy, kinetic and potential. Kinetic energy is the energy associated with motion. An object in motion has the capacity to do work on another object by either pulling it, pushing it, or crashing into it, for examples. Potential energy is the energy associated with 1 This content is available online at .

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position. If by changing position an object can do work on another object, then it has potential energy. For example, a book on a table has potential energy because, if it were to fall to the oor, it could accelerate an object tied to it during the fall. During this study, we will relate chemical energy to these forms of energy. One of the most important principles from Physics is the Conservation of Energy. This tells us that energy is neither created nor destroyed in any process, including a chemical process. Rather, energy is converted from one form to another during these processes. The energy conversion might possibly be from more useful forms to less useful forms of energy, but the energy is nevertheless conserved. We will assume a foundation in the dierent types and energies of chemical bonds. In particular, we must recall that atoms are bonded together when their energy when bonded is lower than their energy when separated. Therefore, breaking a chemical bond requires the input of energy to do work on the bonded atoms by separating them. The more energy required, the stronger the bond.

12.3 Observation 1: Temperature Changes during Chemical Reactions Since we are interested in the energy changes which happen during chemical reactions, it makes sense to look at reactions which have the most conspicuous energy changes, those which evolve heat. Fire is probably the rst known human-controlled chemical reaction. Burning is now understood as a combustion reaction of oxygen with a fuel, such as wood, oil, or natural gas. These reactions were all originally carried out primarily as sources of heat for warmth or cooking. In common terms, we use combustion of fuel to heat up, that is, to make something hotter or, better said, to raise the temperature of something. It is pretty easy to observe that whatever is released during a chemical reaction which makes things hotter is a form of energy. For example, we can carry out the combustion reaction in a closed space that can expand, such as inside a cylinder with a piston inserted to close o the contents of the reaction. As the reaction occurs, we observe that the piston is pushed back, so work is done on the piston, meaning that the reaction has released energy to do that work. (This is the principal mechanism behind an internal combustion engine, of course.) When this transfer of energy creates temperature changes, we call this transfer heat. Since heat can be dened in terms of temperature changes, this tells us that temperature and heat are very closely related concepts. We need a means to measure temperature. It is not enough for us to simply say that something hot has a high temperature. We need a measurement scale that allows us to compare how hot objects are compared to each other. There are lots of ways to do this. All of them are based on measuring some property which correlates to hotness. We most commonly use the expansion and contraction of liquid mercury in a glass tube, but we can observe expansion and contraction of solid metals, gases, etc. Or, we can observe other properties that vary with hotness, like the variation of resistance in wires or thermocouples or like the spectrum of infrared light emitted by a substance. This is why there are so many types of thermometers. As long as they are calibrated against each other so that they give the same reading when the temperature of a specic object is measured, all of them are useful. Once we have a thermometer, we can easily show that heating an object causes its temperature to rise. Perhaps then temperature is the same thing as heat. Let's test this idea and measure the temperature rise produced by a simple heat-producing chemical reaction like burning methane. As an example, we burn 1.0 g of methane gas and use the heat released to raise the temperature of 1.000 kg of water (essentially 1.0 L of water). We observe that the water temperature rises by exactly 13.3 ◦ C. This result is constant for this experiment. By performing this experiment repeatedly, we always nd that the temperature of this quantity of water increases by 13.3 ◦ C. Therefore, the same quantity of heat must always be produced by reaction of this quantity of methane. As such, it is very tempting to say that the amount of heat released by burning 1.0 g of methane is 13.3 ◦ C. If this is true, then every time 1.0 g of methane is burned, a temperature rise of 13.3 ◦ C should be observed. However, if we burn 1.0 g of methane to heat 500 g of water instead, we observe a temperature rise of 26.6 ◦ C. And if we burn 1.0 g of methane to heat 1.000 kg of iron, we observe a temperature rise of 123 ◦ C. Therefore, the temperature rise observed depends on the quantity of material heated as well as what Available for free at Connexions

123 the substance is that is heated. Our temptation has led us astray. 13.3 ◦ C is not an appropriate measure of this quantity of heat, since we cannot say that the burning of 1.0 g of methane "produces 13.3 ◦ C of heat." Such a statement is clearly nonsense, so we must keep the concepts of temperature and heat distinct.

12.4 Observation 2: Heat and Heat Capacity, and Reaction Energy Although temperature and heat are not the same concept, our data do tell us that they are related somehow. Let's look at some additional data. We know that if we burn 1.0 g of methane, the temperature rise in 1.0 kg of water is 13.3 ◦ C or the temperature rise for 0.5 kg of water is 26.6 ◦ C. What if we burn 2.0 g of methane? Experimentally, the temperature rise in 1.0 kg of water is 26.6 ◦ C or the temperature rise for 0.5 kg of water is 53.2 ◦ C. Look at those data carefully. We can reasonably assume that burning twice as much methane generates twice as much heat. And we see that it produces twice the temperature change of a xed amount of water. This tells us that the temperature change for a xed amount of water is proportional to the heat absorbed by the water. Does this work for other materials? Earlier, we used the heat from burning 1.0 g of methane to heat 1.0 kg of iron, and we saw a temperature increase of 123 ◦ C. If we burn 2.0 g of methane to heat 1.0 kg of iron, the temperature increase is found to double to 246 ◦ C. Again, the temperature change is proportional to the heat absorbed. Let's put this in symbols. If we call the quantity of heat q, and ∆T is the temperature rise produced by this heat, then we have observed that q = C ∆T (12.1) where C is a proportionality constant. We need to be careful with this equation, though, because our data say that the relationship between q and ∆T depends on what material is heated (water or iron) and how much is heated (1.0 kg or 0.5 kg). So C depends on these same things: what material is heated and how much of the material is there. C is therefore a property of each material and is called the heat capacity of the material. Our observations have already shown us that the temperature change is double for half as much water. We can repeat these observations for many dierent masses of water, and we also nd that the temperature change is inversely proportional to the mass of the water. This means that the heat capacity C itself is proportional to the mass of the substance heated. (Look back at Equation 1 to convince yourself that this is true. For a xed amount of heat, what happens to the temperature change and the heat capacity if we double the mass of water heated?) So now we rewrite Equation 1 with this new information: q = m Cs ∆T (12.2) Here, m is the mass of the material being heated, and the proportional constant is now called the heat capacity per gram or more commonly the specic heat. Experiments show that, for any particular material, Cs is a relatively constant property of the material. (Cs actually varies slowly with the temperature, so it is about constant unless we make very large temperature changes.) This equation so far is not very helpful, though, because we do not know values for the heat q or for the specic heat Cs . If we knew one, we would know the other from Equation 2, so somehow we have to devise an experiment to measure one or the other. Here's one way to do the experiment. Since heat is a form of energy and energy is the capacity to do work, we just need to measure how much work can be done for a specic amount of heat, e.g. for burning a specic amount of methane. This is tricky, but we've already seen that we can use the heat generated by a reaction to push a piston back in a cylinder. If we burn 1.0 g of methane, we can measure how much work is done on the piston by measuring how much force is generated and for what distance. From these measurements and the rules of physics, we nd that burning 1.0 g of methane can produce a maximum amount of work equal to 55.65 kJ.

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(A second way to do the experiment is to use work to increase the temperature of water and to measure how much work is required to increase the temperature of water by 1 ◦ C. We'll leave it as an exercise to devise a way to elevate temperature by doing work.) What do the data tell us? If 55.65 kJ of work can be done by burning 1.0 g of methane, then burning 1.0 g of methane must produce 55.65 kJ of heat. This is q in Equation 2. But we have already measured that, for 1.0 kg of water, the temperature change is 13.3 ◦ C. This is ∆T in Equation 2, and m is 1000 g. From these data, we can directly calculate that, for water, Cs = 4.184 J/g· ◦ C. This is called the specic heat of water, or somewhat more loosely, the heat capacity of water. Pay attention to the units of this quantity, as they are unusual. In similar ways, it is possible to nd the specic heat or heat capacity of any material of interest. A set of specic heats for dierent substances is shown in Table 12.1: Specic Heat Capacities of Specic Substances at 25 C (unless otherwise noted). This is very valuable for predicting temperature changes in dierent materials. For our purposes, it has an even greater value. We can use this to determine the energy change in a chemical reaction.

Specic Heat Capacities of Specic Substances at 25 C (unless otherwise noted) Substance

Air (sea level) Ar (g) Au (s) CH4 (g) C2 H5 OH (l) CO2 (g) Cu (s) Fe (s) H2 (g) H2 O (0 C) H2 O (25 C) H2 O (100 C) He (g) Ne (g) NaCl (s) O2 (g) Pb (s)

Cp (J/g· ◦C) 1.0035 0.5203 0.129 2.191 2.44 0.839 0.385 0.450 14.30 2.03 4.184 2.080 5.19 1.030 0.864 0.918 0.127

Table 12.1

Calorimetry: Measuring the Heat of a Chemical Reaction Let's illustrate by analyzing the example of burning butane instead of methane. If we burn 1.0 g of butane and allow the heat evolved to warm 1.0 kg of water, we observe that an increase in the temperature of the water of 11.8 ◦ C. Therefore, by (12.2), elevating the temperature of 1000 g of water by 11.8 ◦ C must require 49,520 J = 49.52 kJ of heat. Therefore, burning 1.0 g of butane gas produces exactly 49.52 kJ of heat. Available for free at Connexions

125 The method of measuring reaction energies by capturing the heat evolved in a water bath and measuring the temperature rise produced in that water bath is called calorimetry. Following this procedure, we can straightforwardly measure the heat released or absorbed in any easily performed chemical reaction. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, q > 0 for an endothermic reaction. When heat is evolved, the reaction is exothermic and q < 0 by convention.

12.5 Observation 3: Hess' Law of Reaction Energies The method of calorimetry we have developed works well provided that the reaction is easily carried out in a way that we can capture the energy transfer in a known quantity of water. But many reactions of great interest are very dicult to carry out under such controlled circumstances. Many biologically important chemical reactions may require the conditions and enzymes only available inside a cell. For example, conversion of glucose C6 H12 O6 to lactic acid CH3 CHOHCOOH is one of the primary means of providing energy to cells: C6 H12 O6 → 2 CH3 CHOHCOOH (12.1) Measuring the energy of this reaction is important to understanding the biological process of energy transfer in cells. However, we can't simply put glucose in a beaker and wait for it to turn into lactic acid. Very specic conditions and enzymes are required. We need to develop a dierent method for measuring the energy of this reaction, and this requires more experimentation. To begin our observations, we will work with a few reactions for which we can measure the energy change. Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water: C(s) + 2 H2 O(g) → CO2 (g) + 2 H2 (g) (12.3) Calorimetry reveals that this reaction requires the input of 90.1 kJ of heat for every mole of C(s) consumed. It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that Reaction (3) converts one fuel, C(s), into another, H2 (g). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that C(s) + O2 (g) → CO2 (g) produces 393.5 kJ for one mole of carbon burned; hence q = -393.5 kJ. The reaction

(12.4)

2 H2 (g) + O2 (g) → 2 H2 O(g) (12.5) produces 483.6 kJ for two moles of hydrogen gas burned, so q = -483.6 kJ. Therefore, more energy is available from the combustion of hydrogen fuel than from the combustion of carbon fuel. Because of this, we should not be surprised that conversion of the carbon fuel to hydrogen fuel requires the input of energy. We can stare at the numbers for the heats of these reactions, 90.1 kJ, -393.5 kJ, -483.6 kJ, and notice a surprising fact. The rst number 90.1 kJ, is exactly equal to the dierence between the second number -393.5 kJ and the third number -483.6 kJ. In other words, the heat input in (12.3) is exactly equal to the dierence between the heat evolved in the combustion of carbon ((12.4)) and the heat evolved in the combustion of hydrogen ((12.5)). This might seem like an odd coincidence, but the numbers from the data are too precisely equal for this to be a coincidence. Just as we do anytime we see a fascinating pattern in quantitative data, we should develop a model to understand this. It is interesting that the energy of (12.3) is the dierence between the energies in (12.4) and (12.5), and not the sum. One way to view this is to remember that the energy of a reaction running in reverse must be the negative of the energy of the same reaction running forward. In other words, if we convert reactants to Available for free at Connexions

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CHAPTER 12. MEASURING ENERGY CHANGES IN CHEMICAL REACTIONS

products with some change in energy, and then convert the products back to the reactants, the change in energy of the reverse process must be the negative of the change in energy in the forward process. If this were not true, we could carry out the reaction many times and generate energy, which would violate the Law of Conservation of Energy. Let's list all three reactions together, but in doing so, let's reverse (12.5): C(s) + 2 H2 O(g) → CO2 (g) + 2 H2 (g) C(s) + O2 (g) → CO2 (g) 2 H2 O(g) → 2 H2 (g) + O2 (g) What if we do both (12.4) and (12.8) at the same time?

q = 90.1kJ q = -393.5kJ q = 483.6kJ

(12.6) (12.7) (12.8)

C(s) + O2 (g) + 2 H2 O(g) → CO2 (g) + 2 H2 (g) + O2 (g) (12.9) Since O2 (g) is on both sides of this reaction, it is net neither a reactant nor product, so we could remove it without losing anything. (12.9) is then equivalent to (12.3), except that we carried it out by doing two separate reactions. What if we do (12.9) in two steps, rst doing (12.4) and then doing (12.8)? That would be the same outcome as (12.9) so it must not matter whether we do (12.4) and (12.8) at the same time or one after the other. The reactants and products are the same either way. What would be the energy of doing (12.4) and (12.5) either at the same time or in sequence? It must be the energy of (12.4) plus the energy of (12.5), of course. So if we add these together using the above numbers, the energy of doing both reactions together is -393.5kJ + 483.6kJ = 90.1kJ. This is the energy of (12.9) and it is experimentally exactly the same as the energy of (12.3). This is a new observation. We have found that the energy of a single reaction ((12.3)) is equal to the sum of the energies of two reactions ((12.4) and (12.8)), which together give the same total reaction as the single reaction. When the separate reactions add up to an overall reaction, the energies of the separate reactions add up to the energy of the overall reaction. We could study many reactions in a similar manner to see if this pattern holds up. We nd experimentally that it does. The pattern is therefore a new natural law called Hess' Law. Stated generally (but wordily), the energy of a reaction is equal to the sum of the energies of any set of reactions which, when carried out in total, lead from the same reactants to the same products. This is a powerful observation! (It is important to note here that we have omitted something from our observations. Hess' Law requires that all reactions considered proceed under similar conditions like temperature and pressure: we will consider all reactions to occur at constant pressure.) Why would Hess' Law be true? Why doesn't it matter to the energy whether we carry out a reaction in a single step or in a great many steps which produce the same products? We might have guessed that more steps somehow require more energy or somehow waste more energy. But if so, our guess would be wrong. As such, it would be helpful to develop a model to account for this law to improve our intuition about reaction energies. Figure 12.1 gives us a way we can make progress based on the work we have already done. This is a just a picture of Hess' Law showing all of (12.4), (12.8), and (12.9). (Remember that the total reaction in (12.9) is exactly the same as the original reaction in (12.3).) In Figure 12.1, the reactants C(s) + 2 H2O (g) + O2 (g) are placed together in a box, representing the reactant state of the matter before (12.9) occurs. The products CO2(g) + 2H2(g) + O2 (g) are placed together in a second box representing the product state of the matter after (12.9). The reaction arrow connecting these boxes is labeled with the heat of (12.3) (which is also the heat of (12.9)), since that is the energy absorbed when the matter is transformed chemically from reactants to products in a single step. Also in Figure 12.1, we have added a box in which we place the same matter as in the reactant box but showing instead the products of carrying out (12.4). In other words, we will rst do (12.4), producing CO2 (g) but leaving the H2 O(g) unchanged. Notice that the reaction arrow is labeled with the energy of Available for free at Connexions

127 (12.4). Now we can also add a reaction arrow to connect this box to the product box, because that reaction is just (12.8), producing H2 (g) and O2 (g) from the H2 O(g). And we can label this reaction arrow with the energy of (12.8).

Figure 12.1

This picture of Hess' Law makes it clear that the energy of the reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total energy of the same reaction along the alternative "path" consisting of two steps which connect reactants to products. (This statement is again subject to our restriction that all reactions in the alternative path must occur under the same conditions, e.g. constant pressure conditions.) Now let's take a slightly dierent view of Figure 12.1. Visually make a loop by beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the total energy of reaction as you go. If you go backwards against a reaction arrow, then reverse the sign of the energy, since a reverse reaction has the negative energy of the forward reaction. When we complete a loop and do the sum, we discover that the net energy transferred around the loop starting with reactants and ending with reactants is exactly zero. This makes a lot of sense when we remember the Law of Conservation of Energy: surely we cannot extract any energy from the reactants by a process which simply recreates the reactants. If this were not the case, in other words if the sum didn't equal zero, we could endlessly produce unlimited quantities of energy by following the circuit of reactions which continually reproduce the initial reactants. Experimentally, this never works since energy is conserved. However, we do get a clearer understanding of why adding the reaction energies together gives the total energy of the overall reaction. Hess' Law is a consequence of the Law of Conservation of Energy.

12.6 Using Hess' Law to Measure Reaction Energy It may be hard to remember now, but we started out with observations leading to Hess' Law because we wanted to nd a way to measure the energy of a reaction which can't easily be found using calorimetry. Some reactions require special conditions which are hard to create in a laboratory where we can make measurements. We gave as an example fermentation of glucose to lactic acid: 22 C6 H12 O6 → 2 CH3 CHOHCOOH (12.10) As we noted above, we can't simply put glucose in a beaker and wait for it to turn into lactic acid and measure a temperature change of a water bath. The reaction just doesn't happen without the assistance of enzymes in a cell. Available for free at Connexions

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So let's use Hess' Law, since we know that the energy of (12.10) will be the same as the sum of the energies of any set of reactions which adds up to (12.10). We just need to pick some reactions which are easy to carry out in the laboratory so that we can measure the energies of these reactions. The easiest reactions to conduct, particularly with molecules containing carbon, hydrogen, and oxygen, are almost always combustion reactions. We can pretty easily burn these compounds, reacting them with oxygen to form CO2 (g) and H2 O(g). We can also pretty easily measure the energies of these combustion reactions using calorimetry, just like before. Here is what the experiments give us: C6 H12 O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2 O(g)

(12.11)

2 CH3 CHOHCOOH + 6O2 (g) → 6CO2 (g) + 6H2 O(g) (12.12) We can now follow a two step process which is equivalent to converting one glucose molecule into two lactic acid molecules. First, we burn the glucose, and the energy evolved is -2808 kJ. Second, hypothetically, we convert the CO2 (g) and H2 O(g) into lactic acid and oxygen. Although we can't really do that hypothetical reaction, we don't need to. The energy of that second step is just the negative of the energy we measure for the combustion of two moles of lactic acid, which is -2668 kJ. Using Hess' Law, the overall energy for converting glucose to lactic acid then is just the measured energy of (12.11) plus the negative of the measured energy of (12.12). This is equal to -140 kJ. We now have a means to measure the energy of a reaction which we can actually carry out! This is a fairly general approach, applicable to most materials. By measuring the energies of combustion reactions and then summing those combustion reactions, we can calculate the energy of an overall reaction.

12.7 Review and Discussion Questions 1. How can the temperature of water be elevated by doing work on it? Devise a way to measure the amount work required to raise the temperature of a sample of water by 1 ◦ C. 2. Assume you have two samples of two dierent metals, X and Z. The samples are exactly the same mass. (a) Both samples are heated to the same temperature. Then each sample is placed into separate glasses containing identical quantities of cold water, initially at identical temperatures below that of the metals. The nal temperature of the water containing metal X is greater than the nal temperature of the water containing metal Z. Which of the two metals has the larger heat capacity? Explain your conclusion. (b) If each sample, initially at the same temperature, is heated with exactly 100 J of energy, which sample has the higher nal temperature? 3. Using data from Table 12.1: Specic Heat Capacities of Specic Substances at 25 C (unless otherwise noted), provide two reasons with explanation why a hot object is much more eciently cooled by placing it in water than leaving it in the open air, even when the air and the water are at the same temperature initially. 4. Explain how Hess' Law is a consequence of conservation of energy. 5. The enthalpy of formation of sucrose C6 H12 O6 cannot be measured by the direct reaction of carbon, hydrogen and oxyben. Devise a method to measure ∆Hf for sucrose. What would you measure and how do these measured quantities relate to the ∆Hf for sucrose?

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Chapter 13

Reaction Energy and Bond Energy

1

13.1 Introduction In the previous study, we developed detailed means to observe and measure the energy changes in chemical reactions. This ability is valuable all on its own, since managing the ow of energy from one form to another is a vital economic activity. But our work is not half done. In Chemistry, we seek not just to observe and measure, but also to model and to understand conceptually. We can make this point clearly by thinking about the following. Some chemical reactions produce energy, even in spectacular amounts. The detonation of a single gram of trinitrotoluene (TNT) produces about 4.2 kJ of energy. The reaction of a single gram of sodium metal (Na) with water produces about 8 kJ of energy. On the other hand, some reactions absorb energy, often evidenced by a signicant cooling of the products or the surroundings of the reaction. For example, the hydration of ten grams of ammonium nitrate (NH4 NO3 ) in an instant cold pack absorbs about 3.2 kJ of energy, causing it to be cold enough to treat minor athletic injuries. How can we account for these great variations in the energies of reactions? Where does the energy come from in an exothermic reaction, and where does it in an endothermic reaction? Could we nd a way to predict whether a reaction will be exothermic or endothermic? Answering these questions requires us to develop a model for energy transfer during chemical reactions.

13.2 Foundation We will build signicantly on the results of the previous concept study. We know how to measure energy changes in reactions. A reaction which releases energy into the environment is called an exothermic reaction, and the heat transfer q < 0. A reaction which absorbs energy from the environment is called an endothermic reaction, and the heat transfer q > 0. Hess' Law, developed in the previous concept study, is an extremely important observation. Recall that Hess' Law tells us that the energy of a reaction is equal to the sum of the energies of a set of reactions which add up to the overall reaction. Stated dierently, the energy of a reaction does not depend on what path we follow in converting reactant to products, whether it be in a single reaction or a series of reactions. As long as we start with the same reactants and wind up with the same products, the energy of the reaction is the same. Although this is not an observation or previous conclusion, we'll add to our foundation a denition of a new quantity, called enthalpy. To understand the usefulness of this new quantity, let's remember that, according to Hess' Law, if we start with a set of reactants and carry out a series of reactions which recreate the reactants, then the total energy change summed over that series of reactions has to be exactly zero. Using the Law of Conservation of Energy, this makes sense. We would not expect to be able to change the 1 This content is available online at .

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CHAPTER 13. REACTION ENERGY AND BOND ENERGY

energy of a substance or substances without changing the state of those substances. In fact, for this reason, chemists call the energy of a substance a state function, meaning that the energy depends only on what state the substance is in (gas, liquid, solid; temperature; pressure). An important fact about state functions is the change in a state function from start to nish during a process depends only on the value of the state function at the start and at the nish. The concept of a state function is somewhat analogous to the idea of elevation. The elevation of a city depends only on where the city is located. So, consider the dierence in elevation between Houston and Denver. This dierence is clearly independent of any path we choose to get from Houston to Denver. We could drive a direct route, we could take the train through Chicago, we could take a non-stop or a multi-stop ight, or we could depart Houston on a plane and parachute into Denver. Each path produces exactly the same elevation gain, equal to the elevation in Denver minus the elevation in Houston. Since energy is a state function, the energy of the products depends only on the state of the products and the energy of the reactants only depends on the state of the reactants. So the energy change in going from reactants to products is just the energy of the products minus the energy of the reactants. We could choose any process that takes us from reactant to product. We do not even need to know what process or processes actually happen. You can see how a state function like energy is reected in our observation of Hess' Law. Chemists choose to dene a particular measure of the energy called the enthalpy, designated H. The enthalpy is a state function, just like the energy. The reason that chemists study the enthalpy is that the dierence between the enthalpy of the products, and the enthalpy of the reactants, H(products) - H(reactants) = ∆H, is equal to the heat transfer during a reaction carried out under constant pressure: ∆H = q . Since a great many chemical processes, including essentially all biological processes, happen under constant pressure, this is a very helpful thing to know. For this reason, ∆H is often called the enthalpy of reaction or the heat of reaction. We will measure and calculate ∆H frequently in this concept study.

13.3 Observation 1: Enthalpy of Formation Remember that our task is to understand what determines whether a reaction is exothermic or endothermic and why the energy transfer might be large or small. To do so, we need to get inside the reaction and nd out what energy changes are taking place. And for this, Hess' Law and the state function H are extremely valuable. These allow us to choose any processes we want to carry out a chemical reaction, knowing that the energy transfer doesn't depend on what processes we pick. This frees us to pick processes to observe which are interesting or revealing to us. We want to pick processes which help us compare energies of reactants and products. To do this, it is helpful to think of a standard to compare against. This is like dening elevation relative to sea level, making sea level the standard which is equal to zero elevation. We could have picked the zero to be the highest point on earth, the top of Mt. Everest, or the lowest point on earth, the bottom of the Marianas Trench, or the northernmost point, the North Pole, or anywhere else. We just need to pick something convenient which allows easy comparisons. Once we know every city's elevation relative to sea level, we can easily calculate the elevation change for every possible city to city trip without ever making any of those trips. For the energy of chemical reactions, what standard shall we pick for comparing energies? One way is to pick the set of substances from which we can form all possible other materials. These are, of course, the elements. How would this work? Let's take a specic example. Butene, C4 H8 , reacts with hydrogen to form butane, C4 H10 : C4 H8 + H2 → C4 H10 2 (13.1) To nd the heat of this reaction ∆H, we need to compare the energy of the products to the energy of the reactants. We could do this by subtracting the energy of the products, relative to the energy of the elements, minus the energy of the reactants, again relative to the energy of the same elements. Clearly it would be very helpful to measure these energies. Available for free at Connexions

131 The enthalpy of a substance, say butane, relative to the elements which make it up, carbon and hydrogen, is called the enthalpy of formation of the substance. This is the energy that is either absorbed or released when the substance is made from the elements in their standard form, and is labeled as ∆Hf . For example, for butane, the formation reaction from the elements is 4 C(s) + 5 H2 (g) → C4 H10 (g) (13.2) and the heat of this reaction is called the enthalpy of formation of butane. The enthalpy of formation tells us the energy of butane relative to the elements from which it could be formed. Of course, for this to be useful, we have to measure this quantity. This formation reaction for butane, (13.2), is virtually impossible to perform in a controlled way so that we can measure the energy of the reaction. But, due to the work we did in the previous study on Hess' Law, we have a way to measure the energy of this reaction without actually carrying out the reaction. We can use an alternative pathway. Since carbon, hydrogen, and butane are all easy to burn, we can use combustion reactions as the alternative pathway and measure the energy of these reactions. C4 H10 (g) + 13/2 O2 (g) → 4 CO2 (g) + 5 H2 O(g)

(13.3)

C(s) + O2 (g) → CO2 (g)

(13.4)

H2 (g) + ½ O2 (g) → H2 O(g) (13.5) Using Hess' Law, we can use these three reactions and energies to nd the energy of the formation reaction for butane, in other words, the enthalpy of formation of butane. Figure 13.1 shows a diagram of Hess' Law for (13.2) from which we can write: ∆Hf (C4 H10 )

= ∆H(reaction 3) - 4∆H(reaction 4) - 5∆H(reaction 5) This gives the enthalpy of formation of butane ∆Hf (C4 H10 ) = -125 kJ/mol.

(13.6)

Figure 13.1

Remember that this enthalpy of formation of butane is the energy of (13.2) and therefore measures the energy of butane relative to the elements which make it up. We can use this same approach to observe and measure the enthalpy of formation of any substance we are interested in. The enthalpies of formation of several interesting compounds are listed in Table 13.1: Standard Enthalpies of Formation for Various Substances at 25 C. These values allow us to compare the energies of dierent compounds, since these are all relative to the same standard. Available for free at Connexions

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CHAPTER 13. REACTION ENERGY AND BOND ENERGY

Standard Enthalpies of Formation for Various Substances at 25 C Substance

Acetylene Ammonia Carbon Dioxide Carbon Monoxide Ethanol Ethylene Glucose Hydrogen Chloride Iron(III) Oxide Magnesium Carbonate Methane Nitric Oxide Water (g) Water (l)

Formula

C2 H2 (g) NH3 (g) CO2 (g) CO (g) C2 H5 OH (l) C2 H4 (g) C6 H12 O6 (s) HCl (g) Fe2 O3 (s) MgCO3 (s) CH4 (g) NO (g) H2 O (g) H2 O (l)

∆

H f (kJ/mol)

226.7 -46.1 -393.5 -110.5 -277.7 52.3 -1260 -92.3 -824.2 -1095.8 -74.8 90.2 -241.8 -285.8

Table 13.1

For example, thinking back to (13.1) where butene reacts to become butane, it is interesting to compare the enthalpy of formation of these two compounds to see that the energy of butene is higher than the energy of butane. In fact, we should be able to compare these numbers to measure the energy transfer in (13.1). Hess' Law once again comes in handy as is illustrated in Figure 13.2. Now we can use an alternative pathway leading from reactants to products in two steps, rst from reactants to the elements and then from the elements to the products. This means that we can write that the energy of (13.1) as ∆H

= ∆Hf (C4 H10 ) - ∆Hf (C4 H8 ) (13.7) What about the other reactant in (13.1), H2 ? Why don't we include the enthalpy of H2 in the calculation? Table 13.1: Standard Enthalpies of Formation for Various Substances at 25 C doesn't list an entry for the enthalpy of formation for H2 . This is because the formation reaction for H2 would simply be H2 →H2 , in other words, nothing needs to happen to form H2 from elemental H2 so there is no reaction and no enthalpy of formation for H2 .

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133

Figure 13.2

Our approach for nding ∆Hof (13.1) is perfectly general. We can see that the heat of any reaction can be calculated from ∆H º

= ∆Hf º(products) - ∆Hf º(reactants) (13.8) For this reason, extensive tables of ∆Hfº have been compiled and published. This allows us to calculate with complete condence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be dicult to perform or impossibly slow to react. (13.8) is also a great way to interpret the energies of reactions. For example, if we observe an exothermic reaction, in other words if ∆H 0 spontaneous process in an isolated system It is interesting to consider one particular isolated system, which is the entire universe. Since no energy or matter can transfer in or out of the universe, the universe qualies as isolated. So, another statement of the Second Law of Thermodynamics is: ∆Suniverse > 0 for any spontaneous process Though interesting and quite general, this statement is not yet all that useful because calculating S for the entire universe seems very dicult. In the next Concept Development Study, we will develop a means to do that calculation fairly easily. Available for free at Connexions

CHAPTER 23. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 23.7 Observation 4: Heat Transfer and Entropy Changes 246

We will begin with a simple and common observation about heat ow. Let's take two pieces of metal of the same mass. It doesn't matter what type of metal or what mass, so let's say 100.0 g of copper. Let's heat one of the samples (call it sample A) to 100 ºC and cool the other (call it sample B) to 0 ºC. Then let's place them in contact with each other but insulated from everything surrounded them. We know from everyday experience what is going to happen, but of course, when we run the experiment, the cold metal B warms up, the hot metal A cools o, and both pieces of metal stop changing temperature only when they come to the same temperature, in this case, 50 ºC. We say that they have reached thermal equilibrium. The approach to thermal equilibrium is clearly a spontaneous process: it happens automatically, and indeed there is nothing we can do to stop it other than to isolate the two pieces of metal from each other. Since this is a spontaneous process and since we isolated the two pieces of metal from their surroundings, then the work of the previous study tells us that, in total for system consisting of the two pieces of metal, ∆S > 0. All that has happened is the transfer of heat from one piece of metal to the other. No chemical reactions or phase changes have occurred and there has been no rearrangement of the atoms in the metal samples. Why did this heat transfer increase the entropy? We know from our previous study that raising the temperature of a sample by adding heat increases the entropy of the sample. Let's say that the amount of heat transferred into sample B was q. The energy of sample B goes up by q during the heat transfer. And therefore the entropy of B must go up. However, by conservation of energy, this same amount of heat q must also transfer out of sample A. The energy of sample A goes down by q during the heat transfer. And therefore the entropy of A must go down. Since the heat transfer is the same and the types and masses of the metal samples are the same, it would seem that the entropy increase of B should be exactly equal to the entropy decrease of A, in which case the total entropy change should be zero. But we know that this is not true, because the transfer of heat is a spontaneous process in an isolated system, so ∆S > 0. This can only mean that the entropy increase of the cold sample B must be larger than the entropy decrease of the hot sample A as they approach thermal equilibrium. This tells us that the quantity of heat alone is not enough to predict the entropy change associated with heat transfer. We must also know the temperature to predict the entropy change. Our observations also tell us that the entropy change is greater for the lower temperature sample. Entropy change and temperature are therefore inversely related to each other. Adding heat to a cold sample produces a greater entropy change than adding heat to a hot sample. We can conclude all of this based just upon observing that heat spontaneously ows from a hot piece of metal to a cold piece of metal. What is harder to conclude is what the exact inverse relationship is. With much more eort and mathematical reasoning, we can show that the correct inverse relationship between entropy change and temperature is the simplest inverse relationship, namely an inverse proportion: ∆S = q/T for transfer of heat q at temperature T This equation works when the temperature is held constant during the heating process. If the temperature is not held constant, we have to do an integral of 1/T from the initial temperature to the nal temperature to calculate ∆S. That is not important for our purposes here. But it is interesting to ask how we can heat a body without changing its temperature. We will discuss this in the next section.

23.8 Observation 5: Heat Transfer during Chemical or Physical Processes The results of the previous observations and reasoning are important in all cases where heat is transferred, not just in heat transfers involving substances at two dierent temperatures. Let's think back to our goal here. How can the Second Law be applied to a process in a system that is not isolated? We would like to understand how it is possible for a process in a non-isolated system to be spontaneous when ∆S < 0 for the system in that process. A good example is the freezing of water at temperatures below 0 ºC. What is so Available for free at Connexions

247 special about 0 ºC? Why is it that below this temperature, the freezing process becomes spontaneous even though ∆S < 0? Let's observe the entropy changes for freezing water at -10 ºC, which we know to be a spontaneous process. First, let's calculate the ∆S for the process H2 O(l) → H2 O(s) near 0 ºC. We can do this from a table of absolute entropies, such as in the previous Concept Development Study. At 0 ºC, S for H2 O(l) = 63.3 J/mol·K and S for H2 O(s) = 41.3 J/mol·K, so for the freezing of water, ∆S = -22.0 J/mol·K. As expected, this is less than zero. We know from our previous observation that, since freezing releases heat into the surroundings, then freezing must increase the entropy of the surroundings. Let's calculate this change in entropy, ∆Ssurroundings . We will assume quite reasonably and fairly generally that the only eect on the entropy of the surroundings of the freezing taking place in the system is the transfer of heat from the system to the surroundings. There is no other exchange or interaction which happens. We might expect that releasing heat into the surroundings would change the temperature of the surroundings. However, the surroundings are typically huge (e.g. the entire room in which the process occurs) that the eect of the heat transfer into the surroundings does not make a measurable change in the surrounding temperature. The same is true when energy is absorbed from the surroundings into the system. From our Concept Development Study on energy changes, we can calculate the heat transfer for a process occurring under constant pressure from the enthalpy change for the process, q = ∆Hº. By conservation of energy, the heat ow out of or into the surroundings must be ∆Hº. From our previous observation, the entropy change resulting from this transfer of heat is ∆S ◦ surroundings = q/T = -∆H ◦ /T In this equation, ∆H is the enthalpy change for the system, which is why there is a negative sign in this equation. Notice that, if the process releases energy as is the case in freezing, ∆H ◦ < 0 so ∆S ◦ surroundings > 0. For H2 O(l) → H2 O(s), ∆Hº = ∆Hºf (H2 O(s)) - ∆Hºf (H2 O(l)). From the data tables, we can calculate that ∆Hº = -6.02 kJ/mol. As expected, this is less than zero. From this, we can calculate that at -10 ºC = 263 K, ∆S ◦ surroundings = 22.9 J/mol·K. Again, as expected, this is greater than zero. We now come to the most important observation: the entropy increase of the surroundings is greater than the entropy decrease of the water that is freezing. This means that, in total, the entropy change for both system and surroundings is positive. The conclusion of this observation is that, in analyzing whether a process is spontaneous or not, we must consider both the change in entropy of the system undergoing the process and the eect of the heat released or absorbed during the process on the entropy of the surroundings. In other words, we have to consider the entropy change of the universe: ∆Suniverse = ∆Ssystem + ∆Ssurroundings We can say then that ∆Suniverse = ∆Ssystem + ∆Ssurroundings = ∆Ssystem - ∆Hsystem /T The last part of this equation says that we can calculate the entropy change of the universe from calculations that involve only entropy and energy changes in the system. This sounds very doable! For every spontaneous process, ∆Suniverse > 0, so for every spontaneous process: ∆Ssystem - ∆Hsystem /T > 0 In this inequality, the entropy change and the enthalpy change are both for the system, so the subscripts are now not needed and we can drop them to make the equation easier to read. ∆S - ∆H/T >

0 any spontaneous process

The fact that the temperature appears in this equation is very interesting. Remember why it is there in the denominator: the eect on the entropy of the heat exchange with the surroundings is smaller when the temperature is larger. This also suggests that we try our calculation with a dierent temperature. What if the temperature is 10 ºC, above the freezing point? In this case, we observe experimentally that the melting of solid ice is spontaneous. Let's do the calculation of the inequality above for the process H2 O(l) → H2 O(s) at 10 ºC. As before, ∆Hº = -6.02 kJ/mol and ∆Sº = -22.0 J/mol·K, but now T = 283K. We get that Available for free at Connexions

CHAPTER 23. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

248 ∆S ◦

- ∆H ◦ /T = -0.73 J/mol·K < 0 We nd a negative number, telling us that freezing water is not spontaneous at 10 ºC. More importantly, we know that the reverse process of melting ice is spontaneous at 10 ºC. This means that, if ∆S - ∆H/T < 0, the reverse process is spontaneous. Now we know what is so special about the temperature 0 ºC. Below that temperature, ∆S ◦ - ∆H ◦ /T > 0 for the freezing of water, and above that temperature ∆S ◦ - ∆H ◦ /T < 0 for the freezing of water, so ∆S ◦ - ∆H ◦ /T > 0 for the melting of ice. What happens at 0 ºC? It is easy to calculate that ∆S ◦ - ∆H ◦ /T = 0 at that temperature. Of course, at that temperature at standard pressure, water and ice are in equilibrium. This is a very important observation: when ∆S - ∆H/T = 0, the process is at equilibrium. We will develop this observation in much greater detail in the next Concept Development Study.

23.9 Review and Discussion Questions 1. Each possible sequence of the 52 cards in a deck is equally probable. However, when you shue a deck and then examine the sequence, the deck is never ordered. Explain why in terms of microstates, macrostates, and entropy. 2. Assess the validity of the statement, "In all spontaneous processes, the system moves toward a state of lowest energy." Correct any errors you identify. 3. Recalling the discussion of the freezing point of water, determine what must be true about ∆Hº and ∆Sº for a solid to have a freezing point much higher than that of water. Similarly, determine what must be true about ∆Hº and ∆Sº for a solid to have a freezing point much lower than that of water. 4. Predict the sign of the entropy for the reaction 2 H2 (g) + O2 (g) → 2 H2 O (g). Give an explanation, based on entropy and the Second Law, of why this reaction occurs spontaneously. 5. For the reaction H2(g) → 2 H(g), predict the sign of both ∆Hº and ∆Sº. Based on your knowledge of hydrogen, which of these two terms is probably dominant in determining whether the reaction is spontaneous. What would you predict would be the eect of lowering the temperature? Explain.

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Chapter 24

Free Energy and Thermodynamic Equilibrium 1

24.1 Introduction In the previous study, we observed and applied the Second Law of Thermodynamics to predict when a process will be spontaneous. For example, the melting of solid water at a temperature above 0 C at 1 atm pressure is a spontaneous process, and thermodynamics predicts this very accurately. However, over the course of several Concept Development Studies, we focused on processes at equilibrium, rather than processes occurring spontaneously. These include phase equilibrium, solubility equilibrium, reaction equilibrium, and acid-base equilibrium. Interestingly, we can use our understanding of spontaneous processes to make predictions about equilibrium processes too. To begin, we need to be clear about what we mean by a spontaneous process versus an equilibrium process. At equilibrium, the macroscopic properties we observe (temperature, pressure, partial pressures, concentrations, volume) do not change. We have developed a model to describe equilibrium based on the idea of dynamic equilibrium, meaning that at equilibrium, there are forward and reverse reactions occurring at the molecular level at the same rate. However, this is not what we mean by spontaneous process, since the forward and reverse reactions exactly oset one another in a dynamic equilibrium. By contrast, in a spontaneous process, we observe macroscopic changes: partial pressures of reactants or products are increasing, concentrations are increasing or decreasing, the temperature or volume is changing, etc. This means that the forward and reverse reactions at the molecular level do not oset one another, and real macroscopic changes occur. As we have discovered, during a spontaneous process the entropy of the universe increases. When a process comes to equilibrium, there are no spontaneous processes, so a reaction at equilibrium does not increase the entropy of the universe. We can combine these two ideas to say that, as a process spontaneously approaches equilibrium, the entropy of the universe continually increases until equilibrium is reached, at which point the process no longer increases the entropy of the universe. This gives us a way to predict the conditions under which a process will reach equilibrium. We will develop this approach in this Concept Development Study. We will have to be careful in applying the Second Law of Thermodynamics in calculations. So far, we have only observed and tabulated values of the absolute entropy, S, at standard pressures and concentrations. We can use these to make predictions about processes at standard pressure and concentrations. But we know that phase transitions and reactions almost always come to equilibrium at partial pressures not equal to 1 atm and concentrations not equal to 1 M. Therefore, we must be careful when we interpret calculations of ∆S using S values. And to understand the conditions at equilibrium, we must determine how to calculate S 1 This content is available online at .

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250

values for non-standard conditions. Only then will we be able to apply the Second Law of Thermodynamics at equilibrium conditions.

24.2 Foundation We have come a long way to reach this point in our studies, so we have a substantial foundation to build on. We know all the elements of the Atomic Molecular Theory, including the models for molecular structure and bonding, and we have developed the postulates of the Kinetic Molecular Theory. We have observed and dened phase transitions and phase equilibrium, and we have also observed equilibrium in a variety of reaction systems, including acids and bases. We will assume an understanding of the energetics of chemical reactions, including the idea of a state function and the concept of Hess' Law. In the previous study, we developed the expression ∆S-∆H/T > 0 as a predictor of whether a process will be spontaneous. Chemists generally rewrite this inequality in a dierent and somewhat more convenient form by multiplying both sides by T : -T∆S+∆H < 0 This is the same inequality, but take notice of the less than sign which replaced the greater than sign as a result of multiplying by T. Both terms in this inequality have units of energy, so ∆H - T∆S is often referred to as an energy. However, it is important to remember that this combination of terms arises from calculating the changes in entropy of the system and its surroundings, and the less than sign arises from the Second Law of Thermodynamics which deals with entropy and not energy. As such, chemists refer to this combination of terms as the free energy rather than the energy. Specically, we dene a new term as G = H  TS, which is also called the Gibbs free energy. If the temperature of a process is constant, then for that process ∆G = ∆H - T∆S. This means that ∆G < 0 for any spontaneous process at constant T and P. This might look and sound new and complicated, but this is just the Second Law of Thermodynamics in a succinct and convenient form. To see this, let's recall that ∆Suniverse = ∆Ssystem + ∆Ssurroundings = ∆S - ∆H/T If we multiply this equation by T, we get -T∆Suniverse = -T∆S + ∆H = ∆G These equations show that the change in the free energy is just the change in the entropy of the universe multiplied by T. Since ∆Suniverse > 0, then ∆G 0. The discussion above tells us that, when the freezing of water is at equilibrium, then ∆S ◦ - ∆H ◦ /T = 0. We should be able to solve this equation for T to predict the temperature at 1 atm when liquid water and solid water are in equilibrium. We can calculate ∆Sº = Sº liquid - Sº solid = 22.0 J/mol·K, and ∆Hº = ∆Hº f (liquid) - ∆Hº f (solid) = 6.01 kJ/mol. When we solve for T we get T = 273 K, which is of course the melting point temperature of water. This calculation shows the power of our predictions from the Second Law of Thermodynamics. Not only can we determine what processes are spontaneous, but we also determine the conditions necessary for a process to come to equilibrium! We next want to consider phase equilibrium between liquid and vapor. Our goal is to predict the temperatures and pressures at which equilibrium exists between liquid water and water vapor. Let's start Available for free at Connexions

251 with the temperature at which liquid water and water vapor are at equilibrium when the pressure is 1 atm. We know the observed answer to this; it is the boiling point of water, 100 ºC. Based on our reasoning, ∆S - ∆H/T = 0 at the boiling point temperature at 1 atm pressure. Plugging in ∆Sº = 118.9 J/K·mol and ∆Hº = 44.0 kJ/mol, we can solve for T to nd the equilibrium temperature at 370 K, which is very close to the observed boiling point of 373 K. (The slight error is that the values of ∆Sº and ∆Hº depend somewhat on the temperature, and we used values from data collected at 25 ºC. If we use the accurate values, we can predict the boiling point of water exactly.) At other temperatures and pressures, the study of phase equilibrium involving vapor becomes more complicated because the entropy of a gas depends strongly on the pressure of the gas. This means that we cannot simply use the values of Sº, which are valid at 1 atm pressure. We'll illustrate this by analyzing the thermodynamics of the evaporation of water. At equilibrium, this should lead us to predict the vapor pressure of water and, relatedly, the boiling point of water, but we have some work to do rst. As we recall, the entropy of one mole of vapor is much greater than the entropy of the same amount of liquid. A look back at Table 1 shows that, at 25 ºC, the entropy of one mole of liquid water is 69.9 J/K, whereas the entropy of one mole of water vapor at 1 atm is 188.8 J/K. This means that for the evaporation of one mole of liquid water at room temperature and 1 atm pressure, ∆Sº > 0. We might think that this means that a mole of liquid water at 25 ºC should spontaneously convert into a mole of water vapor, since this process would greatly increase the entropy of the water. We know, of course, that this does not happen. Learning from our previous study, we must also consider the energy associated with evaporation. The conversion of one mole of liquid water into one mole of water vapor results in the absorption of 44.0 kJ of energy from the surroundings. Recall that this loss of energy from the surroundings results in a signicant decrease in the entropy of the surroundings. We combine both of these factors by calculating ∆S - ∆H/T for the evaporation of liquid water at 25 ºC: ∆S - ∆H/T = 118.9 J/K·mol - (44.0 kJ/mol)/(298 K) = - 28.9 J/K·mol We can repeat this calculation in terms of the free energy change: ∆G = ∆H - 4∆S ∆G = 44,000 J/mol  (298.15K)(118.9J/K mol) ∆G = 8.55 kJ/mol > 0 How do we interpret these numbers? Since ∆S ◦ - ∆H ◦ /T < 0 and ∆G > 0, these calculations tell us that the evaporation of water at 25 ºC is not spontaneous. But note carefully that these calculations use the values of entropy and enthalpy changes valid for 1 atm pressure of the water vapor. So, a careful interpretation of these inequalities tells us that evaporation of water at 25 C is not spontaneous when the pressure of the water vapor is 1 atm. We can also conclude from this result that the reverse process is spontaneous: the condensation of water vapor to liquid is spontaneous at 25 C when the pressure of the water vapor is 1 atm. These predictions are all correct and match experimental observations. Although liquid water does have a vapor pressure at 25 C, it is at 23.8 torr, very far below 1.00 atm. So water vapor at 1 atm pressure will condense spontaneously down to a low but non-zero pressure. Now let's determine the pressure at 25 C at which the liquid and vapor are in equilibrium. If we cool the liquid-vapor equilibrium from 100 ºC to 25 ºC while maintaining equilibrium, we observe that the vapor pressure of the water drops to 23.8 torr. This means that, when the temperature is 25 ºC and the pressure is 23.8 torr, ∆G = 0 and ∆S - ∆H/T=0. We have already calculated that 8.55 kJ/mol, and therefore is not zero, so apparently ∆G depends on the pressure. Since ∆G = ∆H  T∆S, then either ∆S or ∆H changes when we change the pressure. Experiments tell us that pressure has little eect on ∆H. This makes sense: in the gas phase, the molecules do not interact, so unless we increase the pressure enormously, changing the separation of the gas particles does not aect their energy. ∆S, it turns out, does depend strongly on pressure. Why does entropy depend so strongly on pressure? From Boyle's Law, we know that 1 mole of water vapor occupies a much larger volume at a low pressure like 23.8 torr than it does at the considerably higher vapor pressure of 1 atm. In that much larger volume, the water molecules have a much larger space to move in, so the number of microstates for the water molecules must be larger in a larger volume. Therefore, the entropy of one mole of water vapor is larger when the volume is larger and the pressure is lower. This Available for free at Connexions

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252

means that ∆S for the evaporation of one mole of water is greater when that one mole evaporates to a lower pressure. We need to calculate how much the entropy of one mole of a gas increases as we decrease the pressure. To do so, we will use our equation S = k ln W. We can reasonably assume that the number of microstates W for the gas molecules is proportional to the volume V, because the larger the volume, the more places there are for the molecules to be. From the previous study, the entropy is given by S = k ln W, which means that S must also be proportional to ln V. Therefore, we can say that S(V2 )  S(V1 ) = R ln W 2  R ln W1 = R ln W2 / W1 = R ln V2 /V1 Since we are interested in the variation of S with pressure rather than volume, we can use Boyle's Law, which states that for a xed temperature, the volume of one mole of gas is inversely proportional to its pressure. This tells us how entropy varies with pressure: S(P2 )  S(P1 ) = R ln P1 /P2 =  R ln P2 /P1 Let's apply this to water vapor. We know that the entropy of one mole of water vapor at 1.00 atm pressure is S = 188.8 J/K. For a pressure of 23.8 torr = 0.0313 atm, Equation (5) shows that S(23.8 torr) = 217.6 J/K for one mole of water vapor. S changes quite a lot as P changes! From this, we can calculate ∆S for the evaporation of one mole of water vapor to 23.8 torr at 25 C: ∆S = S(water vapor at 23.8 torr) - S(liquid water) = 217.6 J/mol·K  69.9 J/mol·K = 147.6 J/mol·K. At equilibrium, ∆S - ∆H/T = 0. When we plug in this value of ∆S and the previous value of ∆H as well as the temperature 298.15 K, we get 0.00 J/mol·K. Or we can calculate ∆G = ∆H - T∆S = 44.0 kJ  (298.15K)(147.6 J/K) = 0.00 kJ. Either of these is the condition we expected for equilibrium. This conrms our observations. We can conclude that the evaporation of water when no vapor is present initially is a spontaneous process with ∆G < 0, and the evaporation continues until the water vapor has reached the equilibrium vapor pressure, at which point ∆G = 0. Equivalently, the condensation of water vapor when the pressure is 1 atm is spontaneous with ∆G < 0 until enough water vapor has condensed to reach the equilibrium vapor pressure, at which point again ∆G = 0. Thermodynamics tells us exactly what the conditions are for phase equilibrium!

24.4 Observation 2: Thermodynamic description of reaction equilibrium Now that we have developed a thermodynamic understanding of phase equilibrium, it is even more useful to examine the thermodynamic description of reaction equilibrium. We want to understand why the reactants and products come to equilibrium at the specic concentrations and pressures that are observed. Recall that ∆G = ∆H  T∆S < 0 for a spontaneous process, and ∆G = ∆H  T∆S = 0 at equilibrium. From these relations, we would predict that most (but not all) exothermic processes with ∆H < 0 are spontaneous, because all such processes increase the entropy of the surroundings when they occur. Similarly, we would predict that most (but not all) processes with ∆S > 0 are spontaneous. We try applying these conclusions to synthesis of ammonia N2 (g) + 3 H2 (g) → 2 NH3 (g) at 298 K, for which we nd that ∆S = 198 J/mol·K. Note that ∆S < 0 because the reaction reduces the total number of gas molecules during the ammonia synthesis, thus reducing W, the number of ways of arranging the atoms in these molecules. ∆S < 0 suggests that N2 and H2 should not react at all to produce NH3 . On the other hand, ∆H = -92.2 kJ/mol from which we can calculate that ∆G = -33.0 kJ/mol at 298 K. This suggests that N2 and H2 should react completely to form NH3 . But in reality, we learned in the Concept Development Study of reaction equilibrium that this reaction comes to equilibrium, with N2 , H2 , and NH3 all present. We determined the equilibrium constant ) = 6 × 105 Kp = P(NP(NH )·P(H ) Why does the reaction come to equilibrium without fully consuming all of the reactants? 3

2

2

2

3

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253 The answer, just as before, lies in a more careful interpretation and application of the values given: ∆S, and ∆G are the values for this reaction at standard conditions, which means that all of the gases in the reactants and products are taken to be at 1 atm pressure. Thus, the fact that ∆G < 0 for Reaction (6) at standard conditions means that, if all three gases are present at 1 atm pressure, the reaction will spontaneously proceed to increase the amount of NH3 . From this analysis, we can say that, since ∆G < 0 for Reaction (6), reaction equilibrium results in the production of more product and less reactant than at standard pressures. This is consistent with our previous measurement that Kp = 6×105 , a number much, much larger than 1. As a result, we can say that Reaction (6) is thermodynamically favorable, meaning that thermodynamics accurately predicts that the equilibrium favors products over reactants. Thermodynamics can even provide a quantitative prediction of the equilibrium constant. Recall that the condition for equilibrium is that ∆G = 0 and that ∆G depends on the pressures of the gases in the reaction mixture, because ∆S depends on these pressures. Above, we showed that the entropy of a gas changes with the pressure by S(P) - S ◦ = -Rln(P/(1 atm)) For the reaction of N2 and H2 to form NH3 , let's use the above equation to calculate ∆S for whatever partial pressures at which the reactants and products happen to be. ∆S = 2SNH3 - 3SH2 - SN2 ∆S = ∆S ◦ - 2Rln(PNH3 ) + 3Rln(PH2 ) + Rln(PN2 ) ∆S = ∆S ◦ - Rln(PNH3 2 /PH2 3 PN2 ) In the last step, we used the properties of logarithms to combine all the terms and to use the coecients as exponents. The term in parentheses looks just like the equilibrium constant expression. However, it is not equal to the equilibrium constant because these partial pressures are not necessarily at equilibrium  they are whatever we choose them to be. As such, we call this term in parentheses the reaction quotient, designated Q : Q = (PNH3 2 /PH2 3 PN2 ) So, ∆S = ∆S ◦ - RlnQ It turns out that this is a general expression for any reaction involving gases, provided that we form Q from the balanced equation with appropriate exponents. It also turns out that this expression works for any reaction involving solutes or ions in solution provided that we take the concentrations of the solutes or ions in solution with the appropriate exponents. We can use this expression to calculate ∆G, if we recall that ∆H does not depend on pressure: ∆G = ∆H - T∆S = ∆H ◦ - T(∆S ◦ - RlnQ) ∆G = ∆H ◦ - T∆S ◦ + RTlnQ Or, nally, ∆G = ∆G ◦ + RTlnQ This last equation is an exceptionally powerful relationship for predicting in what direction a reaction will proceed spontaneously and when the reaction will come to equilibrium. Let's rst consider the reaction at equilibrium. At equilibrium, ∆G = 0. Also Q = K p by denition of Q and K p . So ∆G ◦ = -RTlnKp This is a remarkable equation as it relates a thermodynamic quantity involving enthalpy changes and entropy changes to the experimentally observed reaction equilibrium constant. In fact, this proves why the equilibrium constant we experimentally observed is indeed a constant. Because this result turns out to be general, we can drop the subscript p referring to partial pressures. K is then any equilibrium constant. If ∆Gº is a large negative number, K will be a large positive number and the reaction is thermodynamically favorable. If ∆Gº is a large positive number, K will be a number much less than one, and the reaction is thermodynamically unfavorable. Let's plug the last equation for ∆Gº into the immediately preceding equation for ∆G: ∆G = -RTlnK + RT lnQ = RTln(Q/K) ∆H,

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CHAPTER 24. FREE ENERGY AND THERMODYNAMIC EQUILIBRIUM

We can use this equation to predict when a reaction will be spontaneous from reactants to products, when it will be spontaneous from products to reactants, and when it will be at equilibrium. ∆G will be negative when Q < K, or in other words, when the numerator in Q is too small and the denominator is too large to equal K. This means we need more products and less reactants, so the reaction proceeds spontaneously forward, as we expect when ∆G is negative. Similarly, ∆G will be positive when Q > K, and the reaction proceeds spontaneously in reverse. Only when Q = K will we have ∆G = 0, so equilibrium requires that Q = K. This is what we observed experimentally. Therefore, thermodynamics both predicts and explains the position of equilibrium for chemical reactions. This is an amazing result! Note that the thermodynamic description of equilibrium and the dynamic description of equilibrium are complementary. Both predict the same equilibrium. In general, the thermodynamic arguments give us an understanding of the conditions under which equilibrium occurs, and the dynamic arguments help us understand how the equilibrium conditions are achieved.

24.5 Review and Discussion Questions 1. Why does the entropy of a gas increase as the volume of the gas increases? Why does the entropy decrease as the pressure increases? 2. For each of the following reactions, calculate the values of ∆S, ∆H, and ∆G at T = 298 K and use these to predict whether equilibrium will favor products or reactants at T = 298 K. Also calculate Kp . a. 2CO(g) + O2 (g) → 2CO2 (g) b. O3 (g) + NO(g) → NO2 (g) + O2 (g) c. 2O3 (g) → 3O2 (g) 3. For each of the reactions in Question 3, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants. 4. Show that for a given set of initial partial pressures where Q is larger than Kp , the reaction will spontaneously create more reactants. Also show that if Q is smaller than Kp , the reaction will spontaneously create more products. 5. In our study of phase equilibrium we found that a substance with weaker intermolecular forces has a greater vapor pressure than a substance with strong intermolecular forces. This was explained using dynamic equilibrium arguments. Use thermodynamic equilibrium arguments involving ∆H, ∆S, and ∆G to explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces.

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INDEX

255

Index of Keywords and Terms

Keywords are listed by the section with that keyword (page numbers are in parentheses). Keywords do not necessarily appear in the text of the page. They are merely associated with that section. Ex. apples, Ÿ 1.1 (1) Terms are referenced by the page they appear on. Ex. apples, 1

D E L O

diatomic, Ÿ 11(109), 109 domain, 116 ED, Ÿ 11(109) Electron Domain (ED) Theory, 113 Electron Domain Theory, Ÿ 11(109) expanded valence, Ÿ 11(109), 115 Lewis structure model, Ÿ 11(109), 109 lone pairs, Ÿ 11(109), 112 octahedron, Ÿ 11(109), 116

P T V

polyatomic, Ÿ 11(109), 109 trigonal bipyramid, Ÿ 11(109), 115 trigonal planar, Ÿ 11(109), 116 valence shell electron pair repulsion (VSEPR) theory, 113 valence shell electron pair repulsion theory, Ÿ 11(109) VSEPR, Ÿ 11(109)

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ATTRIBUTIONS

Attributions Collection: Concept Development Studies in Chemistry 2013 Edited by: John S. Hutchinson URL: http://cnx.org/content/col11579/1.1/ License: http://creativecommons.org/licenses/by/3.0/ Module: "Atomic Molecular Theory" By: John S. Hutchinson URL: http://cnx.org/content/m44281/1.6/ Pages: 1-8 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Atomic Masses and Molecular Formulas" By: John S. Hutchinson URL: http://cnx.org/content/m44278/1.6/ Pages: 9-19 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Structure of an Atom" By: John S. Hutchinson URL: http://cnx.org/content/m44315/1.4/ Pages: 21-29 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Electron Shell Model of an Atom" By: John S. Hutchinson URL: http://cnx.org/content/m44287/1.5/ Pages: 31-37 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Quantum Electron Energy Levels in an Atom" By: John S. Hutchinson URL: http://cnx.org/content/m44312/1.4/ Pages: 39-45 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Electron Orbitals and Electron Congurations in Atoms" By: John S. Hutchinson URL: http://cnx.org/content/m44286/1.4/ Pages: 47-60 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/

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257

Module: "Covalent Bonding and Electron Pair Sharing" By: John S. Hutchinson URL: http://cnx.org/content/m44285/1.3/ Pages: 61-69 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Molecular Structures" By: John S. Hutchinson URL: http://cnx.org/content/m44304/1.5/ Pages: 71-83 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/4.0/ Module: "Energy and Polarity of Covalent Chemical Bonds" By: John S. Hutchinson URL: http://cnx.org/content/m44317/1.5/ Pages: 85-95 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Bonding in Metals and Metal-Non-Metal Salts" By: John S. Hutchinson URL: http://cnx.org/content/m44283/1.5/ Pages: 97-108 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Molecular Geometry and Electron Domain Theory" By: John S. Hutchinson URL: http://cnx.org/content/m44296/1.3/ Pages: 109-119 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Based on: Molecular Geometry and Electron Domain Theory By: John S. Hutchinson URL: http://cnx.org/content/m12594/1.1/ Module: "Measuring Energy Changes in Chemical Reactions" By: John S. Hutchinson URL: http://cnx.org/content/m44294/1.4/ Pages: 121-128 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Reaction Energy and Bond Energy" By: John S. Hutchinson URL: http://cnx.org/content/m44306/1.4/ Pages: 129-137 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/

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258

ATTRIBUTIONS

Module: "Physical Properties of Gases" By: John S. Hutchinson URL: http://cnx.org/content/m47710/1.2/ Pages: 139-149 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "The Kinetic Molecular Theory" By: John S. Hutchinson URL: http://cnx.org/content/m47661/1.2/ Pages: 151-159 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Phase Transitions and Phase Equilibrium" By: John S. Hutchinson URL: http://cnx.org/content/m47666/1.2/ Pages: 161-171 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Phase Equilibrium and Intermolecular Forces" By: John S. Hutchinson URL: http://cnx.org/content/m47663/1.2/ Pages: 173-182 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Phase Equilibrium in Solutions" By: John S. Hutchinson URL: http://cnx.org/content/m47664/1.3/ Pages: 183-192 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Reaction Rate Laws" By: John S. Hutchinson URL: http://cnx.org/content/m47744/1.4/ Pages: 193-202 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/4.0/ Module: "Reaction Kinetics" By: John S. Hutchinson URL: http://cnx.org/content/m47740/1.3/ Pages: 203-212 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Reaction Equilibrium in the Gas Phase" By: John S. Hutchinson URL: http://cnx.org/content/m47715/1.2/ Pages: 213-226 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Available for free at Connexions

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259

Module: "Acid Ionization Equilibrium" By: John S. Hutchinson URL: http://cnx.org/content/m47655/1.3/ Pages: 227-237 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/4.0/ Module: "Entropy and the Second Law of Thermodynamics" By: John S. Hutchinson URL: http://cnx.org/content/m47657/1.3/ Pages: 239-248 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/ Module: "Free Energy and Thermodynamic Equilibrium" By: John S. Hutchinson URL: http://cnx.org/content/m47658/1.2/ Pages: 249-254 Copyright: John S. Hutchinson License: http://creativecommons.org/licenses/by/3.0/

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